A Primed Prime

All primes $p>3$ can be expressed in the form $p=6k \pm1$ because $6k \pm 2$ is divisible by $2$ and $6k\pm 3$ is divisible by $3.$

Substituting this into the equation,

$p^{2} = 36k^{2} \pm 12k + 1 \equiv 1 \mod 3$

When $2^{m} \equiv 2 \mod 3$ , then $2^{m+1} \equiv 1 \mod 3$ and $2^{m+2} \equiv 2 \mod 3$

Therefore for all odd n $2^{n} \equiv 2 \mod 3$

Hence for all $p>3$ , $p^{2} + 2^{p} \equiv 0 \mod 3$ and is not prime.

Therefore we only have to check the remaining cases $p=2$ and $p=3$

Clearly only $p=3$ works which gives $p^{2} + 2^{p} = 17$ and we are done :)