I have a solution to this question which I know to be wrong, but I can't find the mistake! The question reads "Let $A$ be a ring in which every element $x$ satisfies $x^n = x$ for some $n>1$. Show that every prime ideal in $A$ is maximal."
My solution goes:
We have $x^n-x = 0 \implies x(x^{n-1}-1) = 0$ so either $x$ is nilpotent or a unit or 0. If $x$ is nilpotent then so is $x^{n-1}$ as the nilradical is an ideal and so by Question 1 we have $x^{n-1}-1$ is a unit in $A$. Thus $$0 = x(x^{n-1}-1)(x^{n-1}-1)^{-1} = x.$$ Hence every non-zero element of $A$ is a unit and then the only prime and maximal ideal is (0).
I can't see where I've made the mistake but I must have somewhere otherwise it implies that all Boolean rings have two elements.
There are two mistakes in one here. It appears you think that you are able to conclude that $x=0$ or $x^{n-1}-1=0$ from this line.
Firstly, as mentioned in a comment, this does not have to be an integral domain, and it need not be true.
Secondly, $x^{n-1}-1=0$ does not imply $x$ is nilpotent, it implies $x$ is a unit.
But basically the reasoning you gave is relevant, because you should be applying all this reasoning to $R/P$ where $P$ is a prime ideal. That proves that $R/P$ is a field, therefore $P$ is maximal.