Let $R$ be a commutative ring with $1\neq 0$. $R$ is said to be von Neumann regular if for all $a\in R$, there is some $x\in R$ such that $a^2x=a.$ Prove that if $R$ is von Neumann regular and $P$ a prime ideal, then $P$ is maximal.
My idea: We know that $P$ is a prime ideal and $R$ is a commutative ring, so $R/P$ is an integral domain. If we can show that $R/P$ is a field, then $P$ is maximal. Further, every finite integral domain is a field, although I am not sure it will be helpful here.
Any suggestions/comments/answers are welcome. Thanks.