Product of nilpotent ideal and simple module is zero

The connection you are looking for is that nilpotent ideals are all contained in the Jacobson radical. This is easy to see since the primitive ideals of a ring are prime, and hence each one has to contain all nilpotent ideals. Thus their intersection (the Jacobson radical) contains all nilpotent ideals.

Since the Jacobson radical annihilates simple $R$ modules, so must each nilpotent ideal.

Let $M$ be a non-zero simple module. As $M$ is simple, $IM = 0$ or $IM = M$. If $IM = M$, then note that $IM = I^{n}M$ $\forall n \geq 1$. But as $I$ is nilpotent, $I^k = 0$ for some $k$. This implies that $M = 0$, a contradiction. Notice that a consequence is that Nilpotent ideals are contained in the Jacobson radical.