A sketch of a direct proof (not using $\,\Bbb Z[i]$ is a UFD). Suppose $\,w = a+bi\,$ and $\,ww' = p\,$ is prime. We show $\,\Bbb Z[i]/w \cong \Bbb Z/p\,$ is a domain, so $\,w\,$ is prime. $ $ Consider the natural image of $\,\Bbb Z\,$ in the quotient ring $\, h:\,\Bbb Z\to \Bbb Z[i]/w\,$ via $\,n\mapsto n+0\,i\pmod{\!w}.\,$ Note $\,p = a^2\!+b^2$ is $\rm\color{#c00}{coprime}$ to $\color{#c00}{a,b}$.
$\ \ h$ is onto: $\ a\!+\!bi\equiv 0 \equiv p\,\Rightarrow\, i\equiv -a/b \,\Rightarrow\, c\!+\!di\equiv c-da/b\ $ (note $\,\color{#c00}{p\nmid b}\,\Rightarrow b^{-1}$ exists mod $p$)
$\ \ker h = p\Bbb Z\:$ by $\,w\mid n \iff p=ww'\!\mid nw'\!=na\!-\!nbi\iff p\mid na,nb\color{#c00}{\iff} p\mid n$
Therefore the First Isomorphism Theorem yields $\ \Bbb Z[i]/w = {\rm im}\ h \,\cong\, \Bbb Z/\ker h = \Bbb Z/p$