Here is one algorithm to factor $a + bi$ when $a \neq 0$, $b \neq 0$:
- Compute the norm of $a + bi$, which is a purely real integer $n$.
- Factor $n$ into $\mathbb{Z}$ primes and if possible further into Gaussian primes.
- From the Gaussian prime factorization of $n$, identify the conjugate pairs. In each conjugate pair, there is one number that belongs in the factorization of $a + bi$ and one that does not. At this point I wish I had something more clever than telling you to try each combination, by dividing $a + bi$ by each number of the conjugate pair and discarding those numbers which don't give Gaussian integers; if both divisions give Gaussian integers, discard one number of the pair arbitrarily.
- If necessary, prefix the factorization with a Gaussian unit (other than 1) to get the signs right.
The most important thing to remember is that the norm function is multiplicative: $N(pq) = N(p) N(q)$. This is something that you can carry over to many real and imaginary rings. Also, if the norm is a number that is prime in $\mathbb{Z}$, that means the corresponding number is prime (or at least irreducible) in the particular domain at hand.
In $\mathbb{Z}[i]$ we have the additional wrinkle that since $d = -1$ the norm function works out to $a^2 + b^2$, which can be occasionally confusing, compared to something more helpful like $a^2 + 2b^2$ in the case of $\mathbb{Z}[\sqrt{-2}]$ or $a^2 - 3b^2$ in the case of $\mathbb{Z}[\sqrt{3}]$.
Review the norm function of $\mathbb{Z}[i]$: $$N(a + bi) = (a - bi)(a + bi) = a^2 + b^2.$$ Thus $$N(-19 + 43i) = (-19 - 43i)(-19 + 43i) = 2210$$ and $2210 = 2 \times 5 \times 13 \times 17$.
Fermat stated and Euler proved that if positive $p = 2$ or $p \equiv 1 \pmod 4$, then $p = a^2 + b^2$. This means that such primes in $\mathbb{Z}$ are composite in $\mathbb{Z}[i]$. The example of $-19 + 43i$ seems to have been contrived specifically to use the first four primes of $\mathbb{Z}^+$ that are composite in $\mathbb{Z}[i]$. We have $$2210 = (1 - i)(1 + i)(1 - 2i)(1 + 2i)(2 - 3i)(2 + 3i)(1 - 4i)(1 + 4i).$$ Since $(-19 - 43i)(-19 + 43i) = 2210$, the factorization of $2210$ overshoots the factorization of $-19 + 43i$ by $-19 - 43i$.
So we try $$\frac{-19 + 43i}{1 - i} = -31 + 12i$$ and $$\frac{-19 + 43i}{1 + i} = 12 + 31i;$$ since both of these give Gaussian integers, I arbitrarily choose to discard $1 + 2i$. Things are more clear-cut with the conjugate pair for 5: $$\frac{-19 + 43i}{1 + 2i} = \frac{67 + 81i}{5}.$$
This leads to $$(1 - i)(1 - 2i)(2 + 3i)(1 + 4i) = 43 + 19i,$$ which is almost correct. We need to swap the real and imaginary parts and change the sign of the new real part. Clearly $-1$ won't do (that would give $-43 - 19i$), and less obviously $-i$ doesn't work either. Then $$i(1 - i)(1 - 2i)(2 + 3i)(1 + 4i) = -19 + 43i.$$
This has been overly laborious and I'm sure someone will come along with a much cleverer way. But even with the algorithm I have presented here I could have skipped Step 4 if I had made a different choice with the conjugate pair for 2, since $$(1 + i)(1 - 2i)(2 + 3i)(1 + 4i) = -19 + 43i.$$