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 actually depends on how you define Noetherian rings. The two most common definitions are:
A ring in which every non-empty set of ideals has a maximal element.
A ring which satisfies the ascending chain condition on ideals — every nondecreasing sequence of ideals is eventually constant.
The fact that (1) implies (2) is clear, but the implication from (2) to (1) is not provable without some weak form of the Axiom of Choice (specifically, the Axiom of Dependent Choice).
Now, using the first definition, it is trivial that every proper ideal $I$ in a Noetherian ring $R$ is contained in a maximal one. Take the collection $\mathcal{A}$ of proper ideals of $R$ that contain $I$. Note that $\mathcal{A}$ is nonempty since $I \in \mathcal{A}$. Therefore, $\mathcal{A}$ has a maximal element $M$, which must be a maximal ideal of $R$ by definition of $\mathcal{A}$.
Using the second definition, it is not provable without any form of choice that every proper ideal $I$ in a Noetherian ring $R$ is contained in a maximal one. It is tempting to proceed by contradiction. Assume there is no maximal ideal that contains $I$. Since $I$ is not maximal, we can find a proper ideal $I' \supsetneq I$. Since $I'$ is not maximal, we can find an ideal $I'' \supsetneq I'$. And so on, thereby obtaining a strictly ascending chain $$I \subsetneq I' \subsetneq I'' \subsetneq \cdots$$ However, each step of this construction requires choosing an ideal extending the previous one, infinitely many choices in total. In the general case, one needs the Axiom of Dependent Choice to justify this.
Note that the Axiom of Dependent Choice is only needed for the general case. For example, if the ring $R$ is countable then the process outlined above can be made effective. Let $x_0,x_1,\dots$ be a fixed enumeration of $R$. To pick $I'$, scan through the enumeration of $R$ until you find an element $x_i$ such that $x_i \notin I$ and $I + Rx_i \neq R$, then let $I' = I + Rx_i$ and continue in the same manner to find $I'', I''', \dots$ If $I$ is not contained in a maximal ideal, then such an $x_i$ can always be found and thus we contradict the ascending chain condition.