It suffices to show for $R$ a positive integer.
Let $P$ be the product of all $z\in \mathbb{Z}[i]$ with $0<|z|<4R$.
Let $D = P + 2R$.
Then if $|z-D|, then $R<|z-P|<3R$. So $z-P$ is a divisor of $P$.
But then $z-P$ divides $z=(z-P) + P$.
On the other hand, since $R<|z-P|$, $z-P$ is not a unit of $\mathbb{Z}[i]$, so $z$ is not a prime.
You have to show additionally that $u(z-P)\neq z$ is not possible for any unit $u\in \mathbb{Z}[i]$, but that's not hard.
As noted above, $|z-P|<4R$.
On the other hand $|z| + R \geq |z| + |z-D| \geq |D| \geq |P| - 2R$. So $|z|\geq |P|-3R$.
Since it is easy to show that $|P|\geq 81R^4>7R$, we see that $|z|>4R$. So $z$ is not a unit times $z-P$.