$p=4n+3$ never has a Decomposition into $2$ Squares, right? Primes of the form $p=4k+1\;$  have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma.
Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic residues $a^2+b^2$ with $a,b\in \Bbb{N}$
$$
a^2 \bmod 4 +b^2 \bmod 4 \le 2?
$$
If so, are there alternate ways to prove it?

Yes.
$a^2 \equiv -b^2 \ \ (p)$.  Then raise to the $\frac{p-1}{2}$ power and use Fermat's little Theorem. You get $p|2$. 
