Solve in prime numbers $$p^q+q^2=2^p$$
My progress:
Clearly by parity, we have either $p^q,q^2$ even or $p^q,q^2$ is odd.
Case 1: $p^q,q^2$ even
Then $q=2,$ so we get $p^2+4=2^p.$ Now, clearly $p=2$ doesn't satisfy. So, we assume $p$ is an odd prime.
Now, we take $\mod 8$ on both sides.
So the $RHS \equiv 0\mod 8$ but the $RHS \equiv 5 \mod 8.$ Not possible.
Case 2: $p^q,q^2$ odd
Now we take $\mod 3$ on both sides. We get $RHS \equiv 2\mod 3.$
Also, $q^2\equiv 0,1 \mod 3$ and $p^q\equiv 0,1,2 \mod 3.$ So we get either $p^q\equiv 2 \mod 3, q^2\equiv 0\mod 3$ or $p^q\equiv 1 \mod 3, q^2\equiv 1\mod 3.$
The first case implies, $p^3+9=2^p.$ Now, for $p\ge 10$ we start getting $ 2^p>>p^3+9.$ Hence, $p\le 10.$ Then putting $p=5$ We get that this doesn't satisfy.
Hence $\boxed{p\equiv 1 \mod 3, 3\nmid q}.$ Hence $p>3.$
We, then take $\mod 8$ on both sides.
We get $RHS\equiv 0\mod 3.$ Also odd squares are $1\mod 8.$ Hence $p^q\equiv 7\mod 8.$
Hence, $\boxed{p\equiv 7\mod 8,~~q^2\equiv 1\mod 8}.$
This was my progress, any hints/solutions?
Thanks in advance.