$x^n\in I$ where $I$ is primary

Question

Let $I$ be primary ideal of a ring $R$ and $\sqrt{I}$ maximal. Suppose $x^n\in I$ for some power $n$.

Is it true that $x\in I$? I do not see any reason to deduce this. As $x^lx^k\in I$ for $l+k=n$. The one can raise $(x^l)^n\in I$ for sure and similarly for $x^k.

Answer 1

This is false. Let $R=\mathbb{Z}$. The ideal $(p^n)$ is primary, and has radical $(p)$, which is maximal. However $p\not\in (p^n)$.