I will show one prime ideal cannot contain another (discounting 0); hence every nonzero prime ideal is maximal. The proof is broken up into several steps.

I use the axiom of choice all over the place. I wonder if a dedekind domain, with chains of prime ideals, might exist, if we set aside the axiom of choice?

If Q lives properly inside a maximal ideal M, M cannot be principal.

A minimal ideal over a generator g is a factor of the ideal {g}.

Suppose another prime ideal lies properly between P and Q. It either contains x or it doesn't. Either way we have a contradiction, since P is minimal over x, and Q is maximal without x. Therefore P and Q are adjacent prime ideals with nothing in between.

Let J be an ideal in RP, and pull back to an ideal H in R. Factor H in R, and push this product forward into RP. Prime ideals remain prime, and only the primes in P persist. Thus J can be factored into prime ideals in the local ring.

Apply this to P when P is a minimal nonzero prime ideal. By correspondence, there is only one nonzero prime ideal in RP, namely f(P). By factorization, every ideal in RP is a product of prime ideals, hence a power of f(P). These ideals are linearly ordered by containment. This means the local ring is a valuation ring. Its valuation group extends forever, hence the powers of the maximal ideal f(P) form a strictly descending chain of ideals.

Let J be the intersection of f(P)n, and suppose J is nonzero. Now J is a power of f(P). Yet the powers of f(P) are strictly descending. This is a contradiction, hence the powers of f(P) intersect in 0.

The powers of P intersect in an ideal H, with f(H) contained in 0. A nonzero ideal embeds in RP, and then extends, thus only 0 has f(0) = 0. Therefore H = 0, and the powers of P intersect in 0.

An ideal such as P7 becomes f(P)7 in the local ring, and then pulls back to P7, or perhaps something larger. But it can't pull back to include P6. If f(P6) = f(P7) then f(P8) = f(P)f(P7) = f(P)f(P6) = f(P7). All images beyond P6 stabilize, and their intersection is f(P6), which is definitely nonzero. This is a contradiction, hence the saturation of each power of P is distinct.

To be more specific, suppose the saturation of P7, call it H, is larger than P7. If the factorization of H includes P7, then H is too small. Thus H is P6 times some other primes outside of P. Push this forward and get f(P)6, not f(P)7. Therefore the saturation of Pn = Pn.

Apply this to the adjacent primes P and Q. Now P becomes a minimal nonzero Prime, and as described above, the powers of P close in on 0. Pull back to R, and the powers of P close in on Q.

What if we localize first, and then mod out by f(Q)? Since Q is saturated, one can mod out by Q, and localize about P, in either order, and get the same ring. So the ideals in RP that contain f(Q) are the ideals of RP mod f(Q), are the ideals in the localization of R/Q, are the powers of f(P). Since they pull back to the powers of P in R/Q, they pull back to the powers of P in R. In other words, the saturation of Pn in R is Pn, whether we mod out by Q or not.

Suppose there is a prime ideal containing G and properly contained in P. Pull these two prime ideals together, so that they are adjacent. Call these adjacent prime ideals S and T, remembering that S might equal P. As shown above, the powers of S intersect in T. This is contained in the powers of P, which intersect in Q. Therefore T is contained in Q. However, T contains g, which is not in Q. This is a contradiction, hence there are no primes between G and P. In other words, P is minimal over G, and is a factor of G.

If P contains another prime ideal T, that is not contained in Q, select g in T-Q, and argue as above to show that T cannot exist. All the prime ideals properly inside P lie inside Q.

Suppose P is the least prime over H. Thus P to some power, times some primes that are not contained in P, but still contain H, equals H. Suppose P contains some other prime Q that is not inside H. We know that Q does not contain H. Select adjacent primes S and T between P and Q. By the above, T does not contain H. The powers of S intersect in T, which includes some x not in H, yet the powers of P cannot contain x outside of H. This is a contradiction, hence P has no other primes inside it, unless those primes live in H.

Let Q be the largest prime in H. (Remember that Q could be 0). Since P and Q are adjacent, the powers of P intersect in Q, and Q = H.

The powers of any prime intersect in another prime.

This completes the circle. Adjacent primes P and Q have Q as the intersection of Pn, and the intersection of Pn is a prime Q, where P and Q are adjacent. (Q might be 0.)

Set Q = 0, and find a unique chain of Prime ideals from P down to 0. This can be done for each maximal ideal.

First, show the local ring is a valuation ring, by showing ideals are linearly ordered. This is the case if the preimages are linearly ordered. The powers of P are linearly ordered, and so are the powers of Q, with Q below all the powers of P. The trick is to show that Q3 times the powers of P all contain Q4. A copy of the chain based at P fits neatly between successive powers of Q.

Let x = yz, where y is a product of 3 elements drawn from Q, and z is another element drawn from Q. Note that yz is in each Q3Pn, and in their intersection. Thus the intersection, which is an ideal, contains Q4. The ideals based on P and Q are well ordered by the exponent on Q followed lexicographically by the exponent on P. A well ordering is a linear ordering, so we're ok. Push this forward and all ideals in the local ring are linearly ordered, hence it is a valuation ring.

The valuation group is an ordered group, and looks the same under any translation. Shift to Q3, and Q3 times the powers of P must increase in valuation, just as Pn increases in valuation. The images in the local ring are distinct, and their preimages are distinct, and factorizations based on P and Q in R correspond uniquely to factorizations in the local ring. Therefore our local ring, with two primes P and Q, is dedekind.

Now for the long sought contradiction. Let g be anything in P-P2. Let g generate G, and factor G. The only prime containing G is P, and P2 misses g, hence G = P. P is principal, and contains Q, which is a contradiction.

Our original hypothesis, one prime ideal containing another, has failed. Every nonzero prime ideal in a dedekind domain is maximal.

If you want a more direct proof, an ideal that will not factor, let H be generated by x2 and y2. If P is a prime factor of H it contains H. Since x times x lies in H, P contains x. Similarly, P contains y, hence P is the maximal ideal K adjoin x and y. Some power of P equals H, and it isn't P1. Since Pn contains xyn-1, which is not in H, H does not factor in R.