You need not resort to any stronger features of $\mathbb{Z}[\mathrm{i}]$ as a ring. Let us consider an arbitrary ring $(A, +, \cdot)$ whose additive subjacent group $(A, +)$ is finitely generated as a $\mathbb{Z}$-module. Then for any $n \in \mathbb{N}^{\times}\colon=\mathbb{N} \setminus \{0\}$ we have that $nA \leqslant_{\mathrm{b}} A$ is a bilateral (the English idiom would be two-sided, but I dislike it) ideal -- fact which is easy enough to ascertain -- and the quotient $A/nA$ is a finite ring.
To show this, let us use the abbreviation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{N}$ and let us also set $B\colon=A/nA$. Remark that $n \in \mathrm{Ann}_{\mathbb{Z}}(B)$, explicitly put $n$ is in the annihilator of the $\mathbb{Z}$-module $A/nA$. This means that this quotient abelian group can be naturally equipped with a $\mathbb{Z}_n$-module structure and it is furthermore easy to see that the lattices of $\mathbb{Z}$-submodules respectively $\mathbb{Z}_n$-modules of $B$ coincide.
Since $B$ is the quotient of a finitely generated $\mathbb{Z}$-module, it remains finitely generated over $\mathbb{Z}$ and therefore -- in light of the above observation regarding the lattices of submodules (with respect to the two distinct rings of operators) -- it is also finitely generated as a $\mathbb{Z}_n$-module. This means there exists a finite set $T \subseteq B$ such that $B$ can be expressed as a quotient (homomorphic image) of $\mathbb{Z}_n^{(T)}=\mathbb{Z}_n^T$, the free $\mathbb{Z}_n$-module on set $T$. As $\mathbb{Z}_n$ and $T$ are both finite, the free module $\mathbb{Z}_n^T$ is also finite and so therefore must also be its quotient $B$.
This general reasoning applies to your particular problem since indeed $\mathbb{Z}[\mathrm{i}]=\langle \{1, \mathrm{i}\}\rangle$ is finitely generated as a(n abelian) group.
P.S. The general argument above only establish finiteness and does little to assess that order of finiteness. In your particular case however, it is clear that the additive group $\mathbb{Z}[\mathrm{i}]$ is a free $\mathbb{Z}$-module of basis $\{1, \mathrm{i}\}$ and therefore isomorphic to $\mathbb{Z} \times \mathbb{Z}$ via the isomorphism:
$$\begin{align}
f \colon \mathbb{Z}\times\mathbb{Z} &\to \mathbb{Z}[\mathrm{i}]\\
f(r,s)&=r+s\mathrm{i},
\end{align}$$
isomorpism which has the property that $f[n(\mathbb{Z} \times \mathbb{Z})]=n\mathbb{Z}[\mathrm{i}]$. Bearing in mind that $n(\mathbb{Z} \times \mathbb{Z})=n\mathbb{Z} \times n\mathbb{Z}$, we thus infer that $f$ induces the following morphism of additive groups:
$$\mathbb{Z}[\mathrm{i}]/n\mathbb{Z}[\mathrm{i}] \approx (\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z}) \approx (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z})=\mathbb{Z}_n \times \mathbb{Z}_n \quad (\mathbf{Gr}),$$
on grounds of which we can claim that $\left(\mathbb{Z}[\mathrm{i}]\colon n\mathbb{Z}[\mathrm{i}]\right)=\left|\mathbb{Z}[\mathrm{i}]/n\mathbb{Z}[\mathrm{i}]\right|=n^2$.
More generally, if ring $A$ had a free $\mathbb{Z}$-module for its subjacent additive group, then the subjacent additive group of the quotient ring $A/nA$ would also be a free $\mathbb{Z}_n$-module such that $\dim_{\mathbb{Z}}A=\dim_{\mathbb{Z}_n}A/nA$.
