Prove that $2^p+p^2$ is prime for $p=3$ only

Whether $p\equiv 1\pmod 3$ or $p\equiv 2\pmod 3$, either way $p^2\equiv 1\pmod 3$

Since $p$ must be odd, $2^p\equiv 2\pmod 3$

Adding them together gets you a number that is divisible by $3$.

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: $$\frac{9k^2 + 12k + 3m + 6}{3} = 3k^2 + 4k + m + 2.$$

Hint: If $p$ is odd, then $2^p+p^2 \equiv -1+p^2 =(p-1)(p+1) \bmod 3$.