Answer: $(p; q) = (2; 7)$ and $(p; q) = (3; 17)$.
If $p = 2$ or $p = 3$, then the obtained quadratic equations with respect to $q$ have two prime solutions: $q = 7$ and $q = 17$.

Now let $p > 3$. We rewrite the initial equation in the form $p^3(p^2 + 1) = (q + 1)(q - 2)$. Note that the greatest common divisor of the numbers $q + 1$ and $q - 2$ is equal to either $1$ or $3$. Since $(q + 1)(q - 2) \div p^3$ and $p \neq 3$ we see that exactly one of the numbers $q + 1$ and $q - 2$ can be divisible by $p$, and so exactly one of them is divisible by $p^3$. If $(q + 1) \div p^3$, then $q + 1 \geq p^3$,

and if $(q - 2) \nmid p^3$, then $q - 2 \ge p^3$. In any case $q \ge p^3 - 1$. Then we obtain $p^5 + p^3 = (q + 1)(q - 2) \ge p^3(p^3 - 3)$, so $p^2 + 1 \ge p^3 - 3$, i.e. $0 \ge p^3 - p^2 - 4 = (p - 2)(p^2 + p + 2)$, which is impossible for $p > 2$. Therefore, there are no solutions different from the solutions mentioned above.