In a pid, an element x is prime iff it is irreducible. One direction is true for any integral domain, so assume x is irreducible. It generates an ideal that is maximal among principal ideals, and since all ideals are principal, it is a maximal ideal. This is also a prime ideal, hence x is prime.
to summarize, x is prime iff x is irreducible, iff x*R is a prime ideal, iff x*R is a maximal ideal, iff R mod x*R is a field.
Every euclidean domain is a pid. Take an ideal H, and consider all the nonzero elements x in H. Select x with the least d(x) in H, and for any y in H, write y = c*x + r. Now r is also in H, and d(r) is suppose to be less than d(x). Since d(x) is minimal, d(r) = 0, r = 0, and y = cx.
If the ring is a noncommutative domain, the above definitions produce a left euclidian domain and a left pid, whereleft ideals are principal, and the former implies the latter via the above proof. We don't encounter this very often.
Returning to integral domains, it is enough to show the prime ideals are principal. That implies a pid.
Start by showing an ascending chain of ideals is never principal. Consider an ascending chain of ever increasing ideals. If x generates the union of these ideals, then x appears in some ideal H in the chain. The next ideal is bigger, and includes elements outside of x*R. Thus x does not generate the union of the chain. The union of any ascending chain of ideals is not principal.
Now consider the ideals of R that are not principal and order them by containment. By the above, the union of a chain of nonprincipal ideals is nonprincipal. Thus zorn's lemma applies, giving a maximal nonprincipal ideal. Such an ideal is known to be prime. But we're assuming all prime ideals are principal, so this is a contradiction. we cannot have a nonprincipal ideal, else it would start the ascending chain; hence R is a pid.