By criteria 2, it is enough to show each prime ideal is principal.
Let H be the sum of any two prime ideals. If this does not span 1 then drive H up to a maximal ideal M, which contains our two prime ideals. Now M contains two different maximal ideals, hence one of them is not maximal after all. This is a contradiction. The prime ideals are pairwise coprime, and if J is the product of these prime ideals, the chinese remainder theorem tells us R is isomorphic to the direct product of fields R/Pi for each prime ideal Pi.
In a generalization of the above, let H be the sum of two prime powers. If this does not span 1 then a maximal ideal M contains both prime powers, and since M is prime, it contains both prime ideals. Once again M contains two maximal ideals, a contradiction. If J is the product of prime powers, R/J is isomorphic to the direct product of the quotient rings R mod these prime powers.
Let P be one of the aforementioned prime ideals and let c lie in P-P2. Apply the chinese remainder theorem to P2, and all other prime ideals. Let g = c mod P2, and 1 mod all other primes. If g lies in some other prime Q then g becomes 0 in the field R/Q, which is a contradiction. Thus g, and the ideal generated by g, does not lie in Q or any power of Q. If Q is part of the unique factorization of g*R, then g would lie in Q. Thus g generates a power of P.
If g were a unit it would be invertible mod P, yet it is 0 mod P. Thus g*R is a positive power of P, at least P1. If g*R is P2 or higher than g becomes 0 mod P2. Yet g = c mod P2, which is nonzero. Therefore g*R = P, and P is primciple. This holds for every prime ideal P, and R is a pid.