Consider the equation modulo $q$, which reduces to $p^3 \equiv p^2 \pmod{q}$. Since $p$ and $q$ are coprime ($p \neq q$ clearly), we must have $p \equiv 1 \pmod{q}$.

Consider now the equation modulo $q^2$. Then the equation becomes $p^3 \equiv p^2 + 2pq \pmod{q^2}$, and since $\gcd(p, q) = 1$, $p^2 \equiv p + 2q \pmod{q^2}$. Let $p = 1 + aq$ for some integer $a$. Then the congruence becomes
$$1 + 2aq \equiv 1 + aq + 2q \pmod{q^2}$$
and we deduce that $a \equiv 2 \pmod{q}$ and hence $p \equiv 2q + 1 \pmod{q^2}$.

We will now that $p < (q+1)^2 = q^2+2q+1$, which will mean that $p = 2q+1$. From the equation we get $p^3 \ge q^5$ and hence $p \ge q$, and hence $(p+q)^2 \le 4p^2$. Hence $p^3 - 4p^2 \le q^5$, or
$$p^2(p - 4) \le q^5.$$
But if $p \ge (q+1)^2$, then $p^2 > q^4$ and $p-4 \ge q^2$ (since $q \ge 2$). We obtain the desired contradiction.

Now substituting $p = 2q + 1$ into the original equation we obtain
$$q^3 - 8q - 3 = 0.$$
The only integer solution is $q = 3$. Hence the only possible solution is $q = 3$, which gives $p = 7$. It is easy to check that this gives a solution to the equation.