Commutative Algebra without the axiom of choice It is well known that in a commutative ring with unit, every proper ideal is contained in a maximal ideal. The proof uses the axiom of choice. This fact, and others that are proved using essentially the same argument, anchor a large part of commutative algebra.
Suppose now that we disallow the use of the axiom of choice. My feeling is that this fact should still hold except for very pathological rings. I would find it odd if commutative algebra was entirely dependent on the axiom of choice. I also recall hearing that there was a workaround argument that did not use the axiom of choice for sufficiently nice rings, so this question is not entirely speculative.
So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done? And what characterizes the rings where we can't?
 A: It should not be surprising at all that the axiom of choice is so involved in most parts of modern mathematics. The reason is simple, too: finitely generated things are by nature very well-behaved, however as time progressed we began exploring things which are not finitely generated (e.g. measure spaces, function rings, etc.) and the axiom of choice is a great tool to control these things.
True, if you are only interested in, say, rings of cardinality less than $2^{2^{\aleph_0}}$ then you probably don't need the entire axiom of choice, small fragments would suffice for everything to be well-behaved. However why limit yourself by cardinality when the argument would hold for all similar structures? So we just assume the axiom of choice and run with it.
What sort of things could break? Here are a few quick examples:


*

*The existence of maximal ideals in unital rings;

*Not every vector space has a basis, and a similar argument would show that not every projective module over $\mathbb Z$ is free;

*For abelian groups injectivity is no longer equivalent to divisibility;

*If things break real bad then there are not enough projective objects in $\mathbf{Ab}$, and not enough injectives either;

*Not every set is an underlying set of a group.


If you get even further to the place where topology begins to take an important part of the work (e.g. topological groups/vector spaces/etc.) then the axiom of choice becomes an even more important tool.
However not all is lost. If your ring can be well-ordered then you can still apply most of the standard arguments to it. If its power set can also be well-ordered then you have even more.
What sort of common sets are guaranteed to be well-ordered in ZF? Well... countable set. That's it. If you assume more, then more. However there is no guarantee to that happening.
So to answer your later question: I am not aware to the existence of such class, furthermore I do not believe there is a "nice" description of rings you can work with in ZF. If anything, I would expect most things to fail at size continuum or less. 
A: As you can see in this related question, the existence of maximal ideals in rings with unity is equivalent to the axiom of choice.
If you are interested in consequence of the axiom of choice, read this.
For rings that have no maximal ideals, read this. Also this paper provides a characterization of commutative rings with no maximal ideals.
EDIT: After seeing the OP's rephrased question in the comments, see this related question.
A: [So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done?]
Most of rings you encounter in algebraic geometry.
For example, rings which are finitely generated over a field and their localizations.
For the proof, see this thread.
