Easily $p=2$ does not work, whereas $(p,n)=(3,4)$ is a solution. Now let $p\ge 5$. We get \[ n(n-1)(n^3+n^2+1) = p^5(p+1). \]Note that $(n,n^3+n^2+1)=(n,n-1)=1$. Likewise, $(n-1,n^3+n^2+1)\in\{1,3\}$. If $p\mid n$ then $p^5\mid n$, which is impossible by size reasons. Likewise if $p\mid n-1$ then $p^5\mid n-1$ as $p>3$, which again is not possible. So, $p^5\mid n^3+n^2+1$. Hence, $n(n-1)\mid p+1$ and $p^5\mid n^3+n^2+1$. From here, we get \[ n(n-1)\le 1+\sqrt[5]{n^3+n^2+1}. \]After some routine calculations, we get $n\in\{1,2\}$, both of which do not work. So, $(3,4)$ is the only solution.