Solution:

Write the given equation in the form
$$p^{2}(p-1)\left(p^{2}+p+1\right)=q(q+1)$$
First observe that it must not be $p=q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct primes, (9) immediately yields $p^{2} \mid q+1$, that is
$$q=a p^{2}-1$$
for some $a \in \mathbb{N}$. Since $p$ and $q$ are both primes, by (9) we get the following cases:

Case 1: $q \mid p-1$, that is
$$p=b q+1$$
for some $b \in \mathbb{N}$. Substituting (11) into (10), and using the fact that $a \geq 1$ and $b \geq 1$, we obtain
$$q=a(b q+1)^{2}-1 \geq(q+1)^{2}-1=q^{2}+2 q$$
a contradiction.

Case 2: $q \mid p^{2}+p+1$, that is
$$p^{2}+p+1=b q$$
for some $b \in \mathbb{N}$. Substituting (10) into (12), we get
$$p^{2}+p+1=b\left(a p^{2}-1\right)$$
If $a \geq 2$, then from (13) it follows that
$$p^{2}+p+1 \geq 2 p^{2}-1$$
or equivalently, $p+1 \geq(p-1)(p+1)$, that is, $(p+1)(2-p) \geq 0$. This implies that $p=2$, and so $q \mid 2^{2}+2+1=7$. Hence, $q=7$, but the pair $p=2$ and $q=7$ does not satisfy the equation (9).

Hence, it must be $a=1$. Then if $b \geq 3$, (13) implies
$$p^{2}+p+1 \geq 3\left(p^{2}-1\right)$$
or equivalently, $4 \geq p(2 p-1)$, which is obviously impossible.

Thus, it must be $a=1$ and $b \in\{1,2\}$. For $a=b=1$, (13) implies that $p=2$, which by (12) again yields $q=7$, which is impossible. Finally, for $a=1$ and $b=2$, (13) gives $p(p-1)=3$, which is clearly not satisfied for any prime $p$.

Hence, there do not exist prime numbers $p$ and $q$ which satisfy the given equation.