This is in follow-up to this question.
Let $R$ be a commutative ring with identity and consider the set $Z \subset R$ of zero divisors. If the ideal $I\subset Z$ is maximal with respect to the constraint, need it be prime?
Since Zorn's Lemma applies equally well to proving that either minimal prime ideals exist or that maximal ideals contained in $Z$ exist, this would provide an alternative proof of that question if true.
A naive approach to proving this is to assume $ab \in I$ with $a, b \notin I$. Since $ab \in I$, it follows that $ab$ is a zero divisor, so $abx = 0$ for some $x \neq 0$, and therefore either $a$ is a zero divisor annihilated by $bx \neq 0$ or $b$ is a zero divisor annihilated by $x$. In either case, the obvious thing to consider would be $aR+I$ or $bR+I$, but there's no obvious reason why either of these ideals should consist of zero divisors.
According to rschwieb, the answer is yes for reduced Noetherian rings.