I want to show that $4 + i$ is an irreducible element of $\mathbb{Z}[i]$.
My current approach is to let $4+i$ = $AB$, where $A$ and $B$ are elements of $\mathbb{Z}[i]$, where $A=a+ia'$ and $B=b+ib'$.
$AB = 4+i$ so:
$$4+i = (ab-a'b') + i(ab' + a'b)$$
therefore: $$ab-a'b' = 4 $$
and: $$ab' + a'b = 1 $$
I don't know where to go from here though. Is this a good start? What are some general strategies for showing something is an irreducible element of a ring?
That is not the way one usually addresses this kind of questions. The key tool is the norm map $N\colon\mathbb Z[i]\to \mathbb Z$ which maps $a+bi\mapsto a^2+b^2$. This has the property that $N(xy)=N(x)N(y)$ and it allows you to translate your question into a question about integers. So if there are $A,B\in\mathbb Z[i]$ s.t. $4+i=AB$ then $17=N(4+i)=N(A)N(B)$, which shows that wlog $N(A)=\pm 1$ and so $A\in\{\pm1,\pm i\}$, proving your claim.
Here's a different approach. Since prime elements are irreducible, it suffices to show that $4+i$ is prime. (In fact, since $\mathbb{Z}[i]$ is a UFD, then an element is irreducible iff it is prime.)
Recall an ideal $I$ of a commutative ring $R$ is prime iff $R/I$ is a domain. Since $\mathbb{Z}[i] = \mathbb{Z}[x]/\langle x^2 + 1 \rangle$, then \begin{align*} \frac{\mathbb{Z}[i]}{\langle 4 + i \rangle} \cong \frac{\mathbb{Z}[x]}{\langle x^2 + 1, x + 4 \rangle} \cong \frac{\mathbb{Z}[x]/\langle x+4 \rangle}{\langle x^2 + 1, x + 4 \rangle/\langle x+4 \rangle} \cong \frac{\mathbb{Z}}{\langle (-4)^2 + 1 \rangle} = \frac{\mathbb{Z}}{\langle 17 \rangle} \end{align*} by the Third Isomorphism Theorem, where the last isomorphism is induced by the evaluation map at $-4$. Since $\mathbb{Z}/\langle 17 \rangle$ is a domain (a field, even), then $\langle 4 + i \rangle$ is prime (actually, maximal). Thus $4+i$ is prime.
