Let A a commutative ring with unity in which every element is idempotent ($x^n=x$ for some n>1 dependent on x), then every prime ideal is maximal. Let $A$ a commutative ring with unity in which every element is idempotent ($x^n=x$ for some n>1 dependent on $x$), then every prime ideal is maximal.
I came across this question revisiting Atiyah's Intro to Commutative Algebra for my comp, and even though it seems very simple I'm stuck. From the theorems I have available, I can only think of the quotient ring route, somehow using idempotence to show that A/p is not only an integral domain, but also a field, so that p is maximal. But I got nowhere with that approach. Any ideas are appreciated.

Your idea is correct. In an integral domain, if $x^n=x$ and $x\neq 0$, then in fact $x^{n-1}=1$ and thus $x$ is invertible. 
