Solution:

The solutions of the equation are $(p, n)=(2,1)$ and $(p, n)=(37,11)$.
We rewrite the equation in the form
$$p(p-1)=(n+1)\left(n^{2}-n+1\right) .$$
We first observe that for every integer $n$ the value of $n^{2}-n+1$ is positive, so all the factors of the equation written above must be positive. We have two cases.

First case: $p \mid n+1$.
For some positive integer $m$ we have $n+1=m p$ and $p-1=m\left(n^{2}-n+1\right)$. But then $n^{2}-n+1 \leq p-1<p \leq n+1$, so $n=1$ and $p=2$. Substituting into the initial equation, one verifies that indeed $(2,1)$ is a solution.

Second case: $p \mid n^{2}-n+1$. For some positive integer $m$ we have $n^{2}-n+1=m p$ and $p-1=m(n+1)$. Substituting the value of $p$ given by the second equation into the first equation, we obtain
$$n^{2}-\left(m^{2}+1\right) n-\left(m^{2}+m-1\right)=0 \text{.}$$
This last equation has integer solutions if and only if its discriminant
$$\Delta=m^{4}+6 m^{2}+4 m-3=\left(m^{2}+3\right)^{2}+(4 m-12)$$
is the square of an integer. This is certainly true if $4 m-12=0$, that is, if $m=3$. Substituting this value, we obtain $n=11$ and $p=37$, a solution of the initial equation ($n=-1$ gives no acceptable solutions).
On the other hand, there are no other values of $m$ for which $\Delta$ is a perfect square. Indeed, it is immediate to see that if $m>3$ then $\left(m^{2}+3\right)^{2}<\Delta<\left(m^{2}+4\right)^{2}$; in the remaining cases, one has by direct substitution that if $m=1$ then $\Delta=8$ and if $m=2$ then $\Delta=21$.
Therefore the solutions of the equation are $(p, n)=(2,1)$ and $(p, n)=(37,11)$.