My first instinct is to use the isomorphism $$\mathbb{Z}[\sqrt{-5}] \cong \mathbb{Z}[X] / (x^2-5)$$ and show that any ideal I in this ring is generated by only two elements. As $\mathbb{Z}[X]$ is noetherian, we can just assume that $$I = (a_{n_1}X^{n_1}+...+a_{1_1}X+a_{0_1},...,a_{n_m}X^{n_m}+...+a_{1_m}X+a_{0_m})$$.
Now I'd like to use that any equivalence class in $\mathbb{Z}[X] / (x^2-5)$ has a representive of degree $\leq1$, so I'd just assume that $$I = (X-a_1,...,X-a_n,b_1,...,b_n)$$ where $a_i, b_i \in \mathbb{Z}$. But how would I proceed from here? Does that help me in any way? Does anyone have a better idea?
Personally, I think that a much simpler approach is to forget about the multiplicative structure for the time being, and think of $\mathbb Z[\sqrt {-5}]$ as an abelian group under addition.
$\mathbb Z[\sqrt {-5}]$ is a free abelian group of rank two. For example, $\{1 , \sqrt{-5}\}$ is a set of generators.
If $I$ is an ideal in $\mathbb Z[\sqrt {-5}]$, then $I$ is also a subgroup of $\mathbb Z[\sqrt{-5}]$. Since $\mathbb Z[\sqrt{-5}]$ is a free abelian group of rank two, $I$ must be a free abelian group of rank at most two, and in particular, one can find a set of at most two elements of $I$ that generates $I$ as an abelian group. These elements will also generate $I$ as an ideal.