Rearrange to get $3^q + 10 = p^3$. Notice that the residues $3^x$ give $\mod{13}$ are $1, 3, 9$ which means the possible $\mod{13}$ values for $p^3$ are $11, 0, 6$ while the cubic residues mod 13 possible are $0, 1, 5, 7, 8, 9, 12$, which means $p^3 \equiv 0\mod{13}$, which means $p = 13$. Substituting this back, we have $q = 7$.

Same as @above

Almost gave up in frustration when I noticed that $\pmod{13}$ worked really well with powers of 3. We claim that the only possible pair is $(p,q)=(13,7)$. It is easy to see that this pair indeed works. We now show that they are the only ones. We first note that the possible residues of $3^n \pmod{13}$ for positive integers $n$ are $1,3$ and $9$. Thus, \[p^3 = 3^q + 10 \in \{11,0,6 \pmod{13}\}\]Also, the possible cubic residues $\pmod{13}$ are simply $0,1,8,-1$ and $-8$. This means, we must have $p^3 \equiv 0 \pmod{13}$. Thus, $13\mid p^3$ which requires $13 \mid p$ which since $p$ is a prime implies that $p=13$ as desired. This upon substitution sets $q=7$ and we are done.

Solved with p^3 (Upwgs_2008)