We have $p^x=(y+1)(y^2-y+1)$. Let $p=2$. Then, $y^2-y+1$ is always odd, forcing $y\in\{0,1\}$ and $x=1$ works. Let $p>2$. If $y=2$ then $p=3$ and $x=2$ works, so assume $y>2$ and in particular $y^2-y+1>y+1$. So $p\mid y+1$ and $p\mid y^2-y+1$ forces $p=3$. Clearly $9\nmid y^2-y+1$ so $y^2-y+1=3$ is the only possibility, forcing $y=2$.