In Dummit & Foote's Abstract Algebra, they make the comment on the fact that unital rings have maximal ideals:
$\dots$ the proof relies on Zorn's Lemma (see Appendix I). In many specific rings, however, the presence of maximal ideals is often obvious, independent of Zorn's Lemma.
Are there specific rings that require use of Zorn's Lemma to show they have maximal ideals?
On
To clarify a subtlety in Henning's answer: no individual ring can possibly require the full strength of Zorn's Lemma. We can always make the axiom of choice fail "very high up" - each set comes with an ordinal rank, and it is always possible to have choice hold for sets of "small enough" rank while failing in general.
That said, the statement "Every ring has a maximal ideal" is equivalent to Zorn's lemma, and as Henning shows for every set $X$ there is a ring $R_X$ definable from $X$ which can't be proved to have a maximal ideal in ZF alone.
Yes:
Let $R$ be the ring of subsets of $\mathbb N$, with the ring addition being symmetric difference and the ring multiplication being intersection. The ring's $0$ is $\varnothing$; $1$ is $\mathbb N$ itself.
Let $\mathfrak F$ be the ideal consisting of all finite subsets of $\mathbb N$.
Does the quotient ring $R/\mathfrak F$ have maximal ideals? Such a maximal ideal would map back to a maximal ideal $\mathfrak M$ of $R$. And then $\{\mathbb N\setminus X\mid X\in\mathfrak M\}$ would be an ultrafilter, which cannot be principal because $\mathfrak F\subseteq \mathfrak M$ contains all singletons.
However, without any form of the Axiom of Choice it is possible that $\mathbb N$ don't have free ultrafilters. (Less than the full AC can do it, though -- but not bare ZF set theory).
Therefore it necessarily requires choice to show that $R/\mathfrak F$ has a maximal ideal.