For $p=2$, $x^2=31$ which is not possible. Let $p$ be odd. Then $7^p-p-16 \equiv (-1)^p-p \equiv -1-p \equiv 0, 1 (\text{mod 4}) \Rightarrow p \equiv 2, 3(\text{mod 4})$. Since $p$ is odd $p \equiv 3(\text{mod 4}) \;\; (1)$ . $x^2 \equiv 7^p-p-16 \equiv -9 (\text{mod p}) \Rightarrow p|x^2+3^2 \; \; \; (2)$. Since $(1)$ and $(2)$ we have $p|9$ which implies $p=3$. For $p=3$, expression equals to $18^2$. Although it is a bit easy for P1, I believe it evaluates usage of techniques which are shown in the summer camps.