= Solution
Let $P=\operatorname{Ann}(m)$ be maximal among annihilators of nonzero elements of $M$. It is a proper <ideal>, since $1m=m\ne0$. Suppose $ab\in P$ and $b\notin P$. Then $bm\ne0$ and
$$
P\subseteq\operatorname{Ann}(bm),\qquad a\in\operatorname{Ann}(bm).
$$
The inclusion holds because every $r\in P$ satisfies $r(bm)=b(rm)=0$. Maximality in the stated set forces $\operatorname{Ann}(bm)=P$, so $a\in P$. This proves the defining implication for a <prime ideal>:
$$
\boxed{ab\in P\ \Longrightarrow\ a\in P\text{ or }b\in P.}
$$
This is a <maximal annihilator of a module element is prime> argument. Maximality among element annihilators is sufficient; $P$ need not be a maximal ideal of $R$.
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