This question have been answered many times but I didn't find a similar proof, I submit it here for verification.
Let $R$ be a Boolean ring (i.e. $\forall r\in R, \exists n>1: r^n=r$). Prove that every prime ideal of $R$ is maximal.
Proof: Let $\mathfrak p\in \operatorname{Spec} R$. Suppose $\exists \mathfrak m\in \operatorname{Spec} R: \mathfrak p\subsetneq \mathfrak m \subsetneq R$.
We work in the quotient space $R/\mathfrak m$ and we denote $\overline x$ the class of $x \in R$.
Let $x\in \mathfrak m \setminus \mathfrak p$.
We have $x^n=x \implies \overline{x}^n=\overline x = \overline 0 $.
Hence $\overline x \in \operatorname{Nilp}(R/\mathfrak m) = \bigcap\limits_{\overline {\mathfrak p} \in \operatorname{Spec}(R/\mathfrak m)}\overline{\mathfrak p} = \bigcap\limits_{ \mathfrak p \in \operatorname{Spec}R} \mathfrak p/\mathfrak m$. This is a contradiction since $x\not\in \mathfrak p$.
Is this proof correct? Thank you for your help, I am suspecting an error in the last line.
Edit
In the last line, the first equality comes from the fact that the nilradical is the intersection of all prime ideals.
The quotient ring over a maximal ideal is a field.
About your proof. It is strange. For example you take $x$ from the maximal ideal and then prove that it is $0$ in the factor-ring. And the last three lines seem to be incorrect. What is the spectrum of a field?