Proving if $2^n + n^2$ is a prime, show that $n ≡ 3 \pmod 6$

Question

If $n$ is a positive integer greater than $1$ such that $2^n + n^2$ is a prime, show that $n ≡ 3 \pmod 6$

Source of the question : http://math.stanford.edu/~paquin/ModPS.pdf

I tried this for hours but couldn't prove it.

My conclusions are

1. $n$ is odd.
2. $2^n + n^2 ≡ 1\pmod 4$

Thank you for everyone in advance!
Pl. help me.

Answer 1

If $n \equiv 0,2,4 \pmod 6$ then that sum is divisible by $2$.

If $n \equiv 1,5 \pmod 6$ then that sum is divisible by $3$.