You've got the right idea. Similar to what you did, from Fermat's little theorem, for some integers $m_1$ and $m_2$, we get that
$$-p \equiv 7 \pmod{q} \;\;\to\;\; p + 7 = m_{1}q \tag{1}\label{eq1A}$$
$$2q \equiv 7 \pmod{p} \;\;\to\;\; 2q = m_{2}p + 7 \tag{2}\label{eq2A}$$
Since $p$ is odd, from \eqref{eq1A}, $p = 3$ gives that $3 + 7 = 2(5) = m_{1}q$. Thus, $q$ only may be $2$ or $5$, with both working. Next, $p = 5$ leads to $5 + 7 = 2^2(3) = m_{1}q$, but neither $q = 2$ nor $q = 3$ works.
As $p = 7$ doesn't work (because then $q = 7$), consider next $p \gt 7$. Since $m_1 \ge 2$ in \eqref{eq1A}, this means $p \gt q$. From \eqref{eq2A}, we thus get that $m_2$ must be a positive odd integer $\lt 3$, so it must be $1$, giving that $2q = p + 7$. This means that $q \gt 7$. From Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger? (or several of its linked questions), we get
$$q \lt p \;\;\to\;\; q^p \gt p^q \tag{3}\label{eq3A}$$
The problem equation then leads to
$$2q^p - p^q = q^p + (q^p - p^q) \gt q^p \;\;\to\;\; 7 \gt q^p \tag{4}\label{eq4A}$$
which is not possible. Thus, $\boxed{(p, q) \in \{(3, 2), (3, 5)\}}$ are the only solutions.