What did I do wrong in my proof of $4\nmid (n^2+3)\implies 2\nmid (n^4-3)$? A homework problem is asking me to prove the following statement:

For any $n\in\mathbb{Z}$, $2\mid (n^4-3)$ if and only if $4\mid (n^2+3)$

Quick note: we're not allowed to use the theory of congruences in our solution.
I figured the best way to approach this was to prove that $4\mid (n^2+3)$ implies $2\mid (n^4-3)$ and $4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$.
The proof of "$4\mid (n^2+3)$ implies $2\mid (n^4-3)$" was not an issue. It's proving "$4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$" where I had some trouble. Here's my argument so far:

If $4\nmid (n^2+3)$, then $n^2+3=4k+r$ for some integer $k$, where $r$ is either $1$, $2$, or $3$. This is equivalent to
$$n^2=4k+r-3$$
so
\begin{align}
n^4 &= (4k+r-3)^2\\
&= 16k^2+8k(r-3)+(r-3)^2\\
\end{align}
which implies that
$$n^4-3=16k^2+8k(r-3)+(r-3)^2-3$$
Since $r$ is either $1$, $2$, or $3$, we have that
$$n^4-3=16k^2+8k(1-3)+(1-3)^2-3=16k^2-16k+1\text{,}$$
$$n^4-3=16k^2+8k(2-3)+(2-3)^2-3=\color{red}{16k^2-8k-2}\text{,}$$
$$\text{or}$$
$$\vdots$$

I stopped here because the part in red is problematic. Factoring out a $2$, it can be seen that $n^4-3$ is even in this case. I can't have this because I'm trying to show that $2\nmid (n^4-3)$, so what did I do wrong? I checked my algebra over and over again, and I can't seem to find a mistake anywhere. I also graphed $(4k+r-3)^2-3$ with $16k^2+8k(r-3)+(r-3)^2-3$ on Desmos for $r=1,2,3$ and the graphs seem identical.
In light of this, it seems like either (1) I'm blind, (2) I made a subtle mistake in my assumptions, or (3) there's something deeper going on that I'm failing to recognize. Am I seeing everything? Are my assumptions for the proof correct? If the answer to both of these is yes, does this mean that dividing $n^2+3$ by $4$ will never yield a remainder $r=2$? If not, what's going on here? Any help is appreciated.

You already noted that if $n^2+3= 4k+2$ then it's a problem. So we should show that $r=2$ is not possible.
Now that $2|n^2+3\implies n$ is odd. But odd squares are of the form $4l+1.$
Hence $n^2+3=4l+4.$
So if $2|n^2+3 $ then $4|n^2+3$ too.
