The answer is no in general.
Take $R = \prod_{n \in \mathbb{N}_{> 0}} \mathbb{Z}/2^n\mathbb{Z}$. Then the Jacobson radical of $R$ is $\prod_{n \in \mathbb{N}_{> 0}} 2\mathbb{Z}/2^n\mathbb{Z}$, which contains a non-nilpotent element. Therefore the Jacobson radical of $R$ doesn't coincide with the nilradical of $R$.
All credits go to Georges Elancwajg, see this MO post.