$x^p-1$

$x^p-1 = \left(x-1\right)\left(x^{p-1}+x^{p-2}+\cdots+x+1\right)$

For $x^p-1$ to be prime, at least one of the two terms in the RHS should become either 0 or 1.

$x^p > 0, \forall x,p>1$ . Therefore, $\left(x^{p-1}+x^{p-2}+\cdots+x+1\right)>1$

$(x-1)=0 \rightarrow x=1$ . But, in this case $x^p - 1 =1^p-1=1-1=0$ , which is not prime. Hence, $x \ne 1$

This leaves only one solution that $x-1=1$ or $\boxed{x=2}$