The following proof was provided/explained to me by @grassmanian on twitter. I thank him again for the support. Unfortunately, neither he nor I have found a way to show that each maximal ideal of contains a unique prime number without using the fact that is a PID, hence the mostly in the title.

We begin with a lemma:

If is an integral domain which has finitely many units, and finitely many maximal ideals, then it is a field.

It is known that any finite integral domain is a field, so we shall assume that is infinite. Consider the Jacobson radical of A, denoted Jac(A). This is defined as follows:

We observe the following: because is an integral domain, we have that for any nonzero , and for any , . is infinite, so if Jac(A) contains any nonzero elements, then will have infinitely many units. This is forbidden by hypothesis, so we are forced to conclude that .

Now, because is an integral domain (in particular because is commutative), we have the following equivalence of definitions:

Where denotes all maximal ideals of .

By hypothesis, has only finitely many maximal ideals. That is, for some . Any two maximal ideals are coprime, so we may conclude the following by the Chinese Remainder Theorem for rings:

Because each is maximal, we have that each is in fact a field, and thus is a product of fields. However, the product of two or more fields is never an integral domain. By hypothesis is an integral domain, so we are forced to conclude that there is only one such , meaning that is indeed a field.

With that lemma out of the way, we proceed.

Because is an integral domain with only finitely many units (), but is not a field, we may conclude that has infinitely many maximal ideals. is a principal ideal domain, so we have that each maximal ideal is of the form for some prime . Hence, there are infinitely many primes.

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