Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:

Prove by induction that $$ \ \forall n \ge 2, \  2^{n+1} > n^{2} + 3$$

My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and   $\  2^{n+1} > n^{2} + 3$
WTS $ \  2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 = n^2 + n^2 + 2n -2n +6 = (n^2+2n+1) + 3 + (n^2 -2n+2) = (n+1)^2 + 3 + (n^2 -2n+2)$.
I am not sure what to do from here. How can I show $ (n^2 -2n+2) > 0?$

$$n^2-2n+2=(n^2-2n+1)+1=\dots$$
