This answer is the same as Zev's, but perhaps stated more "conceptually". For what it's worth:
A commutative ring is von Neumann regular if for all $a \in R$, there is $x \in R$ such that $a^2 x = a$.
Here are two straightforward facts:
Fact 1: Every quotient of a von Neumann regular ring is von Neumann regular.
[The defining condition is an identity, and if an identity holds in a ring it holds in any quotient.]
Fact 2: An integral domain which is von Neumann regular is a field.
[Fact 2 is literally the first thing that springs to mind when I see the somewhat strange defining condition. What does $a^2 x = a$ mean? Well, if we're allowed to cancel the $a$'s, it means $ax = 1$!]
Thus if $\mathfrak{p}$ is a prime ideal in a von Neumann regular ring, $R/\mathfrak{p}$ is a von Neumann regular domain, hence a field, so $\mathfrak{p}$ is maximal.
Best Answer
Yes, your proof is valid, but note that the second implication relies on $R$ being finite. It'd be clearer if written as
The whole thing would be even cleaner if written as