Proving that a Gaussian integer $a+bi$ with $a,b\neq0$ is composite in $\mathbb{Z}[i]$ if its norm $a^2+b^2$ is composite in $\mathbb{N}$?

If you've established certain facts about the possible composite norms (i.e., you can't have primes $3 \pmod 4$ appear an odd number of times), then you can use Euclid's lemma in the Gaussian primes (which you can establish once you define the gcd): $p$ is prime iff when $p$ divides $ab$, it must divide either $a$ or $b$. So if you know that $p = \alpha\overline{\alpha}$ for some Gaussian prime $\alpha$ divides $x^2 + y^2$, you have that $\alpha$ must divide $p$, which divides $x^2 + y^2 = (x + iy)(x - iy)$. Hence, $\alpha$ divides $x + iy$ or $x - iy$. If the former, you're done. If the latter case, choose $\overline{\alpha}$.

That said, this is a lot of annoying casework and I'd rather just follow the usual path to getting a characterization of the Gaussian primes.

Proof by contradiction.

Suppose $x^2+y^2$ is composite in $\mathbb Z$ with non-zero $x$ and $y$. And suppose $x+yi$ is a Gaussian prime.

Let $x^2+y^2=ab$ with $|a|>1$ and $|b|>1$. Then $x+yi$ divides either $a$ or it divides $b$. Without loss of generality assume that $x+yi$ divides $a$. Therefore $x-yi=\frac{a}{x+yi}b$, which must also be a Gaussian prime. This is only possible if $\frac{a}{x+yi}$ is a unit, implying that either $x=0$ or $y=0$. Contradiction.

So if $x^2+y^2$ is composite in $\mathbb Z$ with non-zero $x$ and $y$, then $x+iy$ is not a Gaussian prime.