Well, hopefully a proof right here will suffice instead of some resource. This is a rather nice proof of the case of Gaussian Integers, but it fails to generalize for the other quadratic integer domains without putting in more work at least.
Define the norm $N(a+bi) = a^2 + b^2$. It is straightfoward to check it is multiplicative.
Now, consider a prime $p \equiv 3 \pmod{4}$. Suppose it is not inert, i.e. it factors as $p = \alpha \beta$ for some $\alpha, \beta \in \mathbb{Z}[i]$ and neither of them are units.. Then it is easy to see $N(\alpha) = N(\beta) = p$. However, the equation $a^2 + b^2 = p$ has no solutions modulo $4$, therefore we have reached a contradiction so primes $3 \pmod{4}$ remain inert.
Now to prove primes $1 \pmod{4}$ split. It suffices to show $p = a^2 + b^2$ has a solution where $a,b$ are integers. Let $z$ denote the least value of $\sqrt{-1} \pmod{p}$. Now define $\mathcal L = \{(a,b) \in \mathbb{Z}^2 | a \equiv zb \pmod{p}\}$. It is straightfoward to check $\mathcal L$ is a lattice whose fundamental parallelogram has area $p$. Now by Minkowski's theorem one has $\mathcal L$ contains a nontrivial lattice point inside the circle $x^2 + y^2 < 2p$. Call this point $(a,b)$. But then $a^2 + b^2 \equiv 0 \pmod{p}$ based on the definition of the lattice, thus it must be $a^2 + b^2 = p$. But then $p = (a+bi)(a-bi)$, proving it factors so we are done.
Proving $2$ ramifies is trivial, since it's just $2 = -i(1+i)^2$.
