I know the classification of primes involves remainders $\mod 3$, so what's the error in the reasoning below?
We know $N$ is multiplicative and that $N(\alpha)$ is a unit $\iff\alpha$ is a unit. Let $\alpha$ be irreducible. Writing $\alpha=\beta\gamma$ we know one of $\beta,\gamma$ is a unit, which is equivalent to one of $N(\beta),N(\gamma)$ being units. So the only reason $N(\alpha)$ can be reducible is if there's a presentation $N(\alpha)=a\cdot b$ not coming from an equation $\alpha=\beta\gamma$ in $\mathbb Z[i]$. But how can this be if $\mathbb Z$ is naturally embedded in $\mathbb Z[i]$?
While the codomain of the norm $N$ is $\mathbb N$, its image is comprised of the numbers which are sums of squares. Thus factorizations into products of elements in the image of $N$ do not cover all possible factorizations in $\mathbb N$ and the definition of irreducibility is not met.
As André Nicolas comments, just look at $3$: it is irreducible in $\mathbb Z[i]$ but $N(\alpha)=9=3^2$.