Prove that $ x^4 - 2 $ is irreducible over $ \mathbb{Z}[i] $ 
How do I prove that $ p(x) = x^4 - 2 $ is irreducible over $ \mathbb{Z}[i] $? 

This seems very elementary yet I'm not sure how to do it.
Someone suggested using Eisenstein and $ p = 1+i $, but this doesn't seem right because $ (1+i)^2 = 2i $ is an associate of $ -2 $.
I have seen somewhere that one can use a generalized version of the Rational Root Theorem and simply check that $ 1+i $ and $ 1-i $ are not roots of $ p(x) $, is this correct?
Thank you for your help.

Checking roots is insufficient, as the equation had degree $4$ so it could potentially factored as 2 degree 2 polynomials.
Reduce the equation mod (3+2i). Then $\mathbb{Z}[i]\rightarrow\mathbb{F}_{13}$, $i\rightarrow 5$, and any integer get reduced mod 13=(3+2i)(3-2i). Then the equation is now $X^{4}-2$. If the equation was reducible before, it is still reducible now. But if it is reducible now, either it has a root, or it breaks as product of quadratic factors so either way all of its root lie in $\mathbb{F}_{169}$. Then $\mathbb{F}_{169}^{\times}$ is a cyclic group of order $168$, and $2$ has order $12$, so any roots must have order $48$ which is impossible as $48$ does not divide $168$.
EDIT: same calculation will also work for $\mathbb{F}_{5}$, as the other answer suggested.
