Let's determine those prime numbers $p$ for which $p^{2}+8$ is also prime.
It is known that the square of an integer gives a remainder of 0 or 1 when divided by 3, depending on whether the number itself is divisible by 3 or not.

In the latter case, $p^{2}+8$ is divisible by 3, and since it is greater than 3, it cannot be a prime. Therefore, if $p^{2}+8$ is not divisible by 3, then $p$ must be divisible by 3. The only prime number divisible by 3 is 3, and in this case, $3^{2}+8=17$ is indeed a prime number.

Thus, the problem has only one solution, $p=3$.