Checking if $\langle 2 \rangle$ is a maximal ideal in $\mathbb{Z}[i]$

As Marc indicates, your reasoning is incorrect: $N(z)$ composite does not justify saying that the number $z$ is reducible. Consider Mark's example; $N(3)=3^2$ is composite but $3$ is irreducible.

There are two ways you could go about this problem: factor $2$ (hint: it's associate to a perfect square of a Gaussian integer; find out the absolute value of this Gaussian integer and then go from there), or show the quotient ${\Bbb Z}[i]/(2)\cong{\Bbb Z}[x]/(2,x^2+1)\cong{\Bbb F}_2[x]/(x^2+1)$ is not a field (which means showing that $x^2+1$ is reducible over ${\Bbb F}_2$) since $M\trianglelefteq R$ is maximal $\Leftrightarrow R/M$ is a field.

(It's no coincidence that $2$ is (associate to) a perfect square in ${\Bbb Z}[i]$ and $x^2+1$ is a perfect square in the ring ${\Bbb F}_2[x]$; as a bonus/challenge exercise, find the connection between these two facts.)