$p$ cannot be $2$ since both of the expressions on the right side are odd. $4$ cases. 1. $p\mid 5^{p}-2^{p}$ and $q\mid 7^{q}-2^{q}$. By Fermat's little theorem, $5^{p}-2^{p}\equiv5-2\equiv3\pmod{p}$ and $7-2\equiv0\pmod{q}$. So, $p=3,q=5$. 2. $p\mid 5^{p}-2^{p}$ and $q\mid 5^{p}-2^{p}$. Then $p\mid 5^{p}-2^{p}$, so $p=3$. And $q\mid 5^{3}-2^{3}$ implies $q\in\{3,37\}$. 3. $p\mid 7^{q}-2^{q}$ and $q\mid 7^{q}-2^{q}$. Again, $p\mid 7^{\gcd(q,p-1)}\equiv2^{\gcd(q,p-1)}$. Since $q> p-1$ and $q$ is prime, $\gcd(q,p-1)=1$. So, $p=5$. 4. $p\mid 7^{q}-2^{q}$ and $q\mid 5^{p}-2^{p}$. So, $p\mid 7^{\gcd(q,p-1)}-2^{\gcd(q,p-1)}$ and $q\mid 5^{\gcd(p,q-1)}-2^{\gcd(p,q-1)}$. Again, $\gcd(q,p-1)=1$ so $p=5$. If $\gcd(5,q-1)=1$, then $q\mid 5-2$ which is impossible since $q\geq p$. So, $5\mid q-1$, $q\neq3$ and $q\mid 5^{5}-2^{5}$. So $q$ is any primitive prime divisor of $5^{5}-2^{5}$.