Put $P=(N:M)$.
Because $N$ is a maximal submodule, $P$ is a maximal ideal.
Because $M$ has finite length, there is some minimal $k$ such that $MP^k=MP^{k+1}$. By Nakayama, there exists $s\in 1+P$ such that $MsP^k=0$.
Put $T=MsP^{k-1}$. If $T=0$, then $MP^{k-1}\subset MP^k$, contradicting minimality of $k$. So $T\neq 0$. But by the preceding paragraph, $TP=0$.
So if $k\neq 0$, then any nonzero $m\in T$ will do. But if $k=0$, then $Ms=0$. Therefore $s\in ann(M)\subseteq (N:M)=P$ and hence $1\in P$, a contradiction.