Here I'll use only some of the results that are listed in the body of the question as being known, with the exception that I make implicit use of the result (surely also known?) that irreducibles in $\mathbb{Z}[i]$ are prime.
First, I give two proofs that if $p \equiv 3 \pmod4$ then $p$ is a prime in $\mathbb{Z}[i]$.
This is for no better reason than that I only thought of the shorter proof after writing out the longer one! :(
Both proofs begin with an arbitrary odd prime $p$.
The first proof can of course be skipped without loss.
Stupidly long proof
We know that $p$ is a product of primes in $\mathbb{Z}[i]$.
If $\pi$ is a prime factor of $p$ in $\mathbb{Z}[i]$, then so is $\overline{\pi}$.
If $\overline{\pi} = \pi$, i.e., if $\pi$ is real, then $\pi$ is a factor of $p$ in $\mathbb{Z}$; but $\pi \ne 1$, therefore $\pi = p$.
Therefore, either $p$ is a prime in $\mathbb{Z}[i]$, or else none of the prime factors of $p$ in $\mathbb{Z}[i]$ are real.
If $\pi$ is a non-real factor of $p$ in $\mathbb{Z}[i]$, then $\pi$ and $\overline{\pi}$ cannot be associates, because then they would both be associates of $1 + i$, and $p$ could only be $2$, contradicting the postulate that $p$ is odd.
Therefore, either $p$ is a prime in $\mathbb{Z}[i]$, or else $p$ is a product of the form:
\begin{align*}
\pi_1\pi_2\cdots\pi_n\overline{\pi_1}\overline{\pi_2}\cdots\overline{\pi_n} & = \alpha\overline{\alpha} \\
& = a^2 + b^2 \\
& \equiv 0 \text{ or } 1 \pmod4,
\end{align*}
where $\alpha = \pi_1\pi_2\cdots\pi_n = a + ib$.
Therefore, either $p$ is a prime in $\mathbb{Z}[i]$, or else $p \equiv 1 \pmod4$.
That is, if $p \equiv 3 \pmod4$, then $p$ is a prime in $\mathbb{Z}[i]$. $\square$
(We have not needed to use the non-trivial result that a prime $\equiv 1 \pmod4$ is a sum of two squares.)
More sensible proof
As usual, denote the norm of a Gaussian integer $\alpha = a + ib$ by:
$$
\operatorname{N}(\alpha) = a^2 + b^2 \equiv 0 \text{ or } 1 \pmod4.
$$
If $p = \alpha\beta$ in $\mathbb{Z}[i]$, and neither $\alpha$ nor $\beta$ is a unit, then: $$
p^2 = \operatorname{N}(p) = \operatorname{N}(\alpha)\operatorname{N}(\beta),
$$
where $\operatorname{N}(\alpha) \ne 1$, $\operatorname{N}(\beta) \ne 1$. Therefore, $\operatorname{N}(\alpha) = \operatorname{N}(\beta) = p$. Therefore, $p \equiv 1 \pmod4$.
So, if $p \equiv 3 \pmod4$, then either $\alpha$ nor $\beta$ must be a unit. That is, $p$ is a prime in $\mathbb{Z}[i]$. $\square$
(I use the word "prime" in this way because the word "irreducible" is not used in the question.)
Main result
From now on, we assume that $p \equiv 3 \pmod4$.
Given integers $m$ and $n$, we have:
$$
m + in = p^r\beta,
$$
where $r$ is a non-negative integer, and $\beta$ is a product of primes $\ne p$ in $\mathbb{Z}[i]$.
Therefore, $m - in = p^r\overline{\beta}$, where $\overline{\beta}$ is also a product of primes $\ne p$ in $\mathbb{Z}[i]$.
Therefore, the exponent of $p$ in the prime factorisation of $m^2 + n^2$ in $\mathbb{Z}[i]$ is $2r$.
But if $m^2 + n^2 = p^{2r}h$, where $h = \beta\overline{\beta}$, we cannot have $p | h$ in $\mathbb{Z}$, because then either $p | \beta$ or $p | \overline{\beta}$ in $\mathbb{Z}[i]$, contradicting the definition of $\beta$.
Therefore, the exponent of $p$ in the prime factorisation of $m^2 + n^2$ in $\mathbb{Z}$ is also $2r$. $\square$
