Answer: $(p, q) = (3, 3), (7, 5)$.

If $q = 2$ then no pair $(p, 2)$ satisfies the equation. If $q = 3$ and $q = 5$ then the only pairs satisfying the equation are $(3, 3)$ and $(5, 7)$, respectively.

Let us show that there is no solution for $q \ge 7$. Consider the binary representation of $q$: $q = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_r}$ where $0 \le a_1 < a_2 < \cdots < a_r$ are integers and $r$ is the number of 1's in the binary representation of $q$. For all $1 \le k \le r$ and $1 \le i \le a_k$ the number $\frac{2^{a_k}}{2^i}$ is an integer. Furthermore, when $i > a_k$, we get $\left\lfloor \frac{2^{a_k}}{2^i} \right\rfloor = 0$.

Finally, we have $\sum_{i=1}^{\infty} \left\lfloor \frac{2^{a_k}}{2^i} \right\rfloor = 2^{a_k} - 1$. Therefore, $v_2(q!)$, the highest power of $2$ in $q!$ can be written as
$$v_2(q!) = \sum_{i=1}^{\infty} \left\lfloor \frac{q}{2^i} \right\rfloor = q - r.$$
The original equation is equivalent to $2^{q-2}(2^{p-q+2}-1) = q!$, where $p-q+2 > 0$. Hence $v_2(q!) = q-2$. Therefore, $r=2$ and $q = 2^{a_1}+2^{a_2}$. Since $q$ is a prime number, we get $a_1 = 0$ and $a_2 = 2^t$ for some non-negative integer $t$ ($q$ is a Fermat prime).

As $q \ge 7$ we have $2^{p-q+2} \equiv 1 \pmod 7$ and $p-q+2 \equiv 0 \pmod 3$. Using the fact $q = 2^{2t} + 1 \equiv 2 \pmod 3$ we get $3 \mid p$ and hence $p=3$. For $q \ge 7$ no pair $(3, q)$ satisfies the equation. Done.