I would want someone to go through my attempt instead of suggesting any another answer.
Let $x$ be an nilpotent element which is not present in the prime ideal $P$, and so $x \in a+P$ where $a$ is not zero .
Now $x^m=0$,where $m \in N$. Now, $x^m \in (a+P)^m \in (a^m+P)$. Since $x^m=0$ so $x^m \in P$ and hence $a^m \in P$ .
Since $P$ is a prime ideal so $a.a^{m-1} \in P$ so by induction I can claim that $a \in P$ which is a contradiction to the fact that $a$ is not in $P$.
Your proof is correct, but you take the unnecessary extra step of introducing the element $a$. You could just substitute $x$ for $a$ in the last sentence of your proof.