Let $p$ be a positive integer prime such that $p\equiv1\pmod4$.
Assume that $p$ is a Gaussian prime, so $(p)$ is a maximal ideal of $Z[i]$, and consider the field $K = Z[i]/(p)$.
We state without proof that the set $${a + bi: 0 \le a, b < p}$$ forms a set of coset representatives of $K$ (forgive me for glossing over this detail).
Therefore, $K$ is isomorphic to the finite field $F$ of $p^2$ elements. From here on, we will use $a + bi$ to refer to the coset of $a + bi$ in $Z[i]/(p)$.
Furthermore, the set $A$ containing $$a: 0 \le a < p$$ forms a subfield isomorphic to the finite field of $p$ elements (this is plain from looking at $Z/(p)$, which must be a subring of $K$).
The group of units of $A$ is cyclic (by the primitive root theorem) and contains $4k$ elements, so $A$ contains an element $x$ of multiplicative order $4$.
Of course, the equation $x^4 - 1 = (x^2 + 1)(x^2 - 1) = 0$ implies that $x^2 = -1$, since $A$ contains no $0$ divisors and and $x$ does not have order $2$ (remember that these equations are all modulo $p$).
Therefore, we have $(x^2 + 1) = (x + i)(x - i) = 0$.
Since $x$ is nonzero and $x$ and $i$ are linearly independent in $K$ over $A$, neither $(x + i)$ or $(x - i)$ can be equal to $0$.
Since $K$ contains $0$ divisors, our assumption that $K$ was a field must have been incorrect, so the ideal $(p)$ is not maximal and $p$ is not a Gaussian prime.
I know this proof glosses over several details, but have I made any obvious errors?