The primes in the ring of Gaussian integers $\mathbb Z[i]$ has a simple connection to the ordinary primes: either one of the components is zero and the other is a prime of type $4n+3$ or the norm $\|a+ib\|=a^2+b^2$ is a prime $a\ne 0\ne b$, that is a prime of the type $4n+1$. All $n\in\mathbb Z^+$ with $n=a^2+b^2$, e.g. all ordinary primes of the form $4n+1$, are composites in $\mathbb Z[i]$: $n=(a+ib)(a-ib)$.
There is essentially one pair of Gaussian primes such that $|z_1-z_2|=1$, and that is $(1+i,2+i)$. If defining prime pairs in $\mathbb Z[i]$ with $|z_1-z_2|=2$, then it seems like for each integer $n>0$ there is a prime $n+im\in\mathbb Z[i]$ with a prime as a pair. Can this be proved?
Verified for all $a+ib\in\mathbb Z[i]$ with $0<a,b\leq1000$.
If $|z_1-z_2|=2$ one can assume e.g. that $\Re (z_1-z_2)=0$ and $|\Im (z_1-z_2)|=2$. Hence, the conjecture can be formulated as:
For each $a\in\mathbb N^+$ it exists $b\in\mathbb Z$ such that $a+ib$ and $a+i(b+2)$ are Gaussian primes.
Equivalent conjecture: $\forall n\in\mathbb N^+\exists m\in\mathbb Z: n^2+m^2,n^2+(m+2)^2\in\mathbb P$.