Proofs-Divisibility of Integers

Let $\displaystyle n \in \mathbb{Z}$. Prove that $\displaystyle 2\ |\ (n^4 - 3) $ if and only if $\displaystyle 4\ |\ (n^2 + 3) $.

Attempt at Proof:

First I just dealt with the forward implication: if $\displaystyle 2 \ |\ (n^4 - 3) $ then $\displaystyle 4 \ |\ (n^2 + 3) $. To prove this I took the contrapositive of this statement which is: If $\displaystyle 4 \not\vert\ (n^2 + 3) $, then $\displaystyle 2 \not\vert\ (n^4 - 3) $.

$\displaystyle 4 \not\vert\ (n^2 + 3) $ can be broken down into 3 cases, the case I'm having trouble with is: let $\displaystyle (n^2 + 3) = 4q + 2$, then $\displaystyle n^2 = 4q - 1 $ and then $\displaystyle n^4 - 3 = (4q - 1)^2 - 3 = 16q^2 - 8q - 2 = 2(8q^2 -4q - 1) $ which is an even number and so $\displaystyle 2 \ |\ (n^4 - 3) $, but that's not what I wanted for the contrapositive.

What am I doing wrong?

Thanks in advance.