Let $A$ be commutative ring with $1$. Suppose $I$ is an ideal of $A$. Define $(I:x)=\{y\in A : yx\in I\}$. Then $(I:x)$ is an ideal and it is called colon ideal.
Consider set of all ideals of such form. Then it forms a poset under natural order.
My question is -
if $(J:x)$ is maximal element of a chain in the poset then it is prime ideal?
I tried proving by contradiction. If $ab\in (J:x)$ then $abx\in J$. Suppose that neither $ax\in J$ nor $bx\in J$. Then consider an ideal $K$ in $A$ such that $a\in K$. Now look at $(K:x)$ then I want to somehow contradict the maximality.
Let $R$ be a commutative unital ring.
$(I: x)=\operatorname{Ann}_{R}(\bar x)$ where $\bar x\in R/I$ and $R/I$ is seen as an $R$-module. Of course this is $R$ if $\bar x=0$ i.e. $x\in I$. So, as is pointed out in the comments, you probably should impose $x\notin I$.
One can then ask if $M$ is an $R$-module and $P$ is a maximal element of the set $\Sigma=\left\{\operatorname{Ann}_R m:m\neq 0\right \}$, is $P$ prime. Indeed it is. Let $P=\operatorname{Ann}_R m_0$. Say $ab\in P$. Then $a\notin P\implies am_0\neq 0\implies\operatorname{Ann}_R m_0\subseteq \operatorname{Ann}_R am_0\in \Sigma $ since $am_0\neq 0$. The maximality forces $\operatorname{Ann}_R m_0=\operatorname{Ann}_R am_0\ni b\implies b\in P$.
Hopefully this answers your question to some extent.