<s>[INST] Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.
 [/INST] So you know that all primes larger than 3 are of the form $3k + 1$ or $3k + 2$.
Do you know which of those forms $2^p$ is? Obviously $2 = 3 \times 0 + 2$. Then $4 = 3 \times 1 + 1$, $8 = 3 \times 2 + 2$, $16 = 3 \times 5 + 1$, $32 = 3 \times 10 + 2$, etc. In short, $2^n \equiv 2 \pmod 3$ if $n$ is odd, so if $p$ is a prime number greater than 2, it's odd, and consequently $2^p \equiv 2 \pmod 3$. Let's say $2^p = 3m + 2$ for later reference.
Now let's review what happens with the squares of primes. Suppose $p = 3k + 2$. Then $p^2 = (3k + 2)^2 = 9k^2 + 12k + 4 \equiv 1 \pmod 3$.
And so $2^p + p^2 = (3m + 2) + (9k^2 + 12k + 4) = 9k^2 + 12k + 3m + 6$, and, just in case it's not obvious enough: $$\frac{9k^2 + 12k + 3m + 6}{3} = 3k^2 + 4k + m + 2.$$
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