I just want to make sure my proof is alright, I would appreciate some hints for a direct proof also. What I have doesn't feel very elegant.
Assume for contradiction that $\alpha, \beta$ are zero-divisors in $\mathbb{Z}[i]$. Then, $$ (ac-bd)+(ad+bc)i=0 $$ To be zero both $\operatorname{Re}(\alpha\beta)=0$ and $\operatorname{Im}(\alpha\beta)=0$, so $ac-bd=0$ and $bc+ad=0$. Then $ac=bd$ and $bc=-ad$, multiplying both sides of $ac=bd$ by $c$ and substituting in for $bc$ gives that $acc = -add$, but this gives $c^2=-d^2$ which is a contradiction.