Definition 1
Let $A$ be a (not necessarily commutative) ring.
Let $M$ be a left $A$-module.
Suppose $M \neq 0$ and $M$ has no proper $A$-submodule other than $0$.
Then we say $M$ is simple.
Definition 2
Let $A$ be a (not necessarily commutative) ring.
Let $M$ be a left $A$-module.
Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a finite descending sequence of $A$-submodules.
If each $M_i/M_{i+1}, i = 0, 1, ..., n - 1$, is simple, this sequence is called a composition series of $M$.
The $n$ is called the length of the composition series.
We define the length of the $0$-module as $0$.
Definition 3
Let $A$ be a (not necessarily commutative) ring.
Let $M$ be a left $A$-module.
Suppose $M$ has a composition series, the lengths of each series are the same by Jordan-Holder theorem. We denote it by $leng_A M$ or $leng$ $M$.
If $M$ does not have a composition series, we define $leng_A M = \infty$.
Lemma 1
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Let $M_1 \supset M_2$ be $A$-submodules of $M$.
Suppose $M_1/M_2$ is simple.
Let $N$ be an $A$-submodule of M.
Then $(M_1 + N)/(M_2 + N)$ is simple or $0$.
Proof:
Let $f:M_1 \rightarrow (M_1 + N)/(M_2 + N)$ be the restriction of the canonical morphism
$M_1 + N \rightarrow (M_1 + N)/(M_2 + N)$.
Since $f(M_2) = 0$, $f$ induces a morphism $g:M_1/M_2 \rightarrow (M_1 + N)/(M_2 + N)$.
Since $f$ is surjective, $g$ is also surjective.
Since $M_1/M_2$ is simple, the assertion follows.
QED
Lemma 2
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Suppose $n = leng$ $M$ is finite.
Let $N$ be an $A$-submodule of $M$.
Then $leng$ $M/N \leq n$.
Proof:
This follows immediately from Lemma 1.
Lemma 3
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Suppose $n = leng$ $M$ is finite.
Let $N$ be a proper $A$-sumodule of $M$.
Then $leng$ $N \leq n - 1$.
Proof:
We use induction on $n$.
If $n = 0$, the assertion is trivial.
Hence we assume $n \geq 1$.
Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a composition series.
Since $leng$ $M_1 = n - 1$, if $N \subset M_1$, $leng$ $N \leq leng$ $M_1$ by the induction assumption.
Hence we can assume that $M_1 \neq N + M_1$.
Since $M_1 \subset N + M_1 \subset M$, $M = N + M_1$.
Since $N/(N \cap M_1)$ is isomorphic to $M/M_1$, it is simple.
Hence it suffices to prove that $leng$ $N \cap M_1 \leq n - 2$.
Suppose $N \cap M_1 = M_1$.
Then $N = M_1$ or $N = M$.
Since $N$ is a proper submodule, $N = M_1$.
But this contradicts our assumption.
Hence $N \cap M_1 \neq M_1$.
By the induction assumption, $leng$ $N \cap M_1 \leq n - 2$.
QED
Lemma 4
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Suppose $leng$ $M$ is finite.
Let $N_1 \subset N_2$ be $A$-submodules of $M$.
Suppose $leng$ $N_1 = leng$ $N_2$.
Then $N_1 = N_2$.
Proof:
$leng$ $N_2 = leng$ $N_1 + leng$ $N_2/N_1$.
Hence $leng$ $N_2/N_1 = 0$.
Hence $N_1 = N_2$.
QED
Lemma 5
Let $A$ be a commutative ring.
Suppose $leng$ $A$ is finite.
Let $\Lambda$ be nonempty set of ideals of $A$.
Then there exists a maximal element in $\Lambda$.
Proof:
Let $r = sup$ {$leng$ $I$; $I \in \Lambda$}.
Since $r$ is finite, there exists $I \in \Lambda$ such that $r = leng$ $I$.
By Lemma 4, $I$ is a maximal element of $\Lambda$.
QED
Lemma 6
Let $A$ be a commutative ring.
Suppose $leng$ $A$ is finite.
Let $\Lambda$ be nonempty set of ideals of $A$.
Then there exist a minimal element in $\Lambda$.
Proof:
Let $r = inf$ {$leng$ $I$; $I \in \Lambda$}.
Since $r$ is finite, there exists $I \in \Lambda$ such that $r = leng$ $I$.
By Lemma 4, $I$ is a minimal element of $\Lambda$.
QED
Lemma 7
Let $A$ be a commutative ring.
Suppose $leng$ $A$ is finite.
Then every ideal of $A$ is finitely generated.
Proof:
Let $I$ be an ideal of $A$.
Let $\Lambda$ be the set of finitely generated ideals contained $I$.
Since $0 \in \Lambda$, $\Lambda$ is not empty.
By Lemma 5, there exitst a maximal element $I_0$ in $\Lambda$.
Suppose $I \neq I_0$.
There exists $x \in I - I_0$.
Let $I_1$ be the ideal generated by $I_0 \cup$ {$x$}.
Since $I_1 \in \Lambda$ and $I_1 \neq I_0$, this is a contradiction.
Hence $I = I_0$.
QED
Lemma 8
Let $A$ be a commutative ring.
Suppose $leng$ $A$ is finite.
Let $J$ be the intersection of all the maximal ideals of $A$.
Then there exists an integer $k \geq 1$ such that $J^k = J^{2k}$.
Proof:
Let $\Lambda$ = {$J^k; k = 1, 2, ...$}.
By Lemma 6, there exists a minimal $J^k$ in $\Lambda$.
Since $J^k = J^{k+1} = ...$, $J^k = J^{2k}$.
QED
Proposition
Let $A$ be a commutative ring.
Suppose $leng$ $A$ is finite.
Let $J$ be the intersection of all the maximal ideals of $A$.
Then $J$ is nilpotent.
Proof:
By Lemma 8, there exists an integer $k \geq 1$ such that $J^k = J^{2k}$.
Let $I = J^k$.
Then $I = I^2$.
By Lemma 7, I is finitely generated.
By Nakayama's lemma, there exists $r \in A$, such that $r \equiv 1$ (mod $I$) and $rI = 0$.
We claim that r is invertible.
If otherwise, $rA$ is a proper ideal.
By Lemma 5, there exists a maximal ideal $P$ such that $rA \subset P$.
Since $I \subset P$, $0 \equiv 1$ (mod $P$).
This is a contradiction.
Hence $r$ is invertible.
Since $rI = 0$, $I = 0$ as desired.
QED
The axiom of choice is in fact equivalent to the assertion that every commutative unital ring has a maximal ideal.
Since the negation of the axiom of choice is as non-constructive as the axiom of choice itself, we can only say that there exists a commutative unital ring without a maximal ideal when the axiom of choice fails. The sets which are involved in this process are non well-orderable.
It is important to understand that much like the axiom of choice only assures that certain objects exist; its negation has a very similar intangible action: it only assures that some sets cannot be well-ordered, some partial orders in which every chain is bounded will not have maximal elements, and some families of non-empty sets do not have a choice function. We have no means of knowing where these objects are without further assumptions.
It is consistent that the axiom of choice holds for every set you have ever dreamed using, but then fails acutely. In that universe you cannot imagine to actually find that set which cannot be well-ordered, or that ring without a maximal ideal, and so on. But it is also consistent that the axiom of choice fails "nearby" and the counterexamples appear in relatively familiar sets (objects related to, or defined from the real numbers for example).
The best way to actually "find" a ring without a maximal ideal is to follow the proof of how the existence of maximal ideals imply the axiom of choice. These proofs often consists of taking a family of non-empty sets, defining some ring (or whatever) and using the maximal ideal to prove the existence of a choice function. Therefore starting with a family of non-empty sets which does not have a choice function guarantees that the process fails and that the ring defined in such proof will not have any maximal ideals.
Added:
Some point that came up in my comment exchange with Trevor Wilson under his answer, is that the axioms are syntactical. They allow us to write proofs. It seems to me, upon re-reading this question that you think something like:
Take a universe of ZFC, $R$ is a unital ring and $I$ is a maximal ideal. Now just don't assume that AC holds. Now we can't prove that $I$ is a maximal ideal.
That's false for two main reasons:
Once you fixed a universe of sets, the axiom of choice is either true or false in that universe. Even if you don't assume it, it has a truth value, and since we began from a universe of ZFC this truth value is indeed true. We just might not be able to write a proof from ZF that $I$ is maximal, but this is something that is still true in that universe of sets.
So changing the assumptions does not necessarily mean that you have changed your universe of sets.
If you have a definition for a ring (e.g. $\Bbb{R^R}$ with pointwise addition and multiplication) then the actual underlying set, and more importantly its subsets, may change between one universe of set theory and another. So for example we cannot prove from ZF that $\Bbb{R^R}$ has a maximal ideal because there are universes of set theory where this is false. But the underlying set $\Bbb{R^R}$ is very different between those universes, and again more importantly, its power set is different.
So to sum up all my answer (with its two additions), simply removing the assumption that the axiom of choice holds will not falsify the axiom of choice. It might be that you just won't be able to write a proof that there exists a maximal ideal in every unital ring.
If, however, you allow yourself to change the universe of set theory then it is possible that a certain ring will "lose" its maximal ideals, simply because we removed those sets from the universe (and in some cases did a whole revamp of the universe altogether).
Best Answer
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:
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.