Versions of Nakayma's Lemma
Solution 1:
Since $a\in I$, we know in particular that $a$ belongs to the unique maximal ideal $m\subseteq R$. Clearly, $1+a$ cannot belong to $m$, as otherwise we would have $1=(1+a)-a\in m$, a contradiction. Thus, $1+a$ is invertible, and so $(1+a)M=0$ implies $M=0$.
More generally, if $R$ is any ring, $M$ a finitely generated module, and there is an ideal $I$ which is contained in the Jacobson radical of $R$ and satisfies $IM=M$, then $M=0$.