This question comes from Examples 17.6.3 and 41.1.2 in John Voight's Quaternion Algebras.
The two exercises both work with the definite quaternion algebra $B := \left(\frac{-1,-23}{\mathbb{Q}}\right)$, and the maximal order $\mathcal{O} := \mathbb{Z} + \mathbb{Z}i + \mathbb{Z}\frac{1 + j}{2} + \mathbb{Z}i\frac{1 + j}{2}$. The overall goal is to compute the $2$-Brandt matrix $T(2)$.
In Example 17.6.3, the author produces a list of 3 right $\mathcal{O}$-ideals, and it seems to be implied that these are all of the right $\mathcal{O}$-ideals of reduced norm $2$. I don't follow the reasoning for why this list is the complete list of such ideals. In the paragraphs below, I've reproduced what I hope is enough of the background for someone with more knowledge than me to fill in the details of the argument.
The author sets the notation $\alpha := i$ and $\beta := \frac{1 + j}{2}$. We now note that since $B$ is split at $2$, we have an embedding $\mathcal{O} \hookrightarrow M_2(\mathbb{Z}_2)$. By considering the minimal polynomials of $\alpha$ and $\beta$, we can see that under this embedding, $\alpha$ maps to $\begin{pmatrix}0 & -1 \\\ 1 & 0 \end{pmatrix}$ and $\beta$ maps to $\begin{pmatrix}1 & 0 \\\ 0 & b_0\end{pmatrix}$, where $b_0$ satisfies $b_0^2 - b_0 + 6 = 0$ and $b_0 \equiv 0 \pmod{2}$.
The author then notes that $$\beta, \beta + 1, (\alpha + 1)\beta \equiv \begin{pmatrix}1 & 0\\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\\ 0 & 1\end{pmatrix},\begin{pmatrix}1 & 0\\\ 1 & 0\end{pmatrix} \pmod{2},$$ and considers the ideals $$I_{(1:0)} = 2\mathcal{O} + \beta \mathcal{O}, \quad I_{(0:1)} = 2\mathcal{O} + (\beta - 1)\mathcal{O}, \quad I_{(1:1)} = 2\mathcal{O} + (\alpha + 1)\beta\mathcal{O}.$$
These are supposedly 1.) right ideals of reduced norm 2 in $\mathcal{O}$, and 2.) the only such ideals. Neither fact is entirely clear to me, but there is a reference to Lemma 17.6.1, which makes me think that is the key ingredient. That lemma says the following:
Let $e \in \mathbb{Z}_{\geq 0}$. Then every principal right $M_2(\mathbb{Z}_p)$-ideal $I$ with $\operatorname{nrd}(I) = p^e$ is of the form $I = \alpha M_2(\mathbb{Z}_p)$ where $$\alpha \in \left\{ \begin{pmatrix}p^u & 0 \\\ c & p^v\end{pmatrix} : u, v \in \mathbb{Z}_{\geq 0}, u + v = e, \text{ and } c \in \mathbb{Z}/p^v\mathbb{Z} \right\}.$$
This is an important point, so good that you are asking if you are confused! It's indeed explained in Lemma 17.6.1 which makes a forward reference to Lemma 26.4.1. I'll try to summarize the key points in this special case.
First of all, an ideal can be understood locally: that's the local-global dictionary, Theorem 9.1.1. So there is a bijection between right ideals of $\mathcal{O}$ of norm $2$ and right ideals of $\mathcal{O}_2$ of norm $2$. The upshot is that $\mathcal{O}_2 \simeq \mathrm{M}_2(\mathbb{Z}_2)$, and all right ideals of a matrix ring are principal. This means we need to write down the elements of $\mathrm{M}_2(\mathbb{Z}_2)$ up to right multiplication by $\mathrm{GL}_2(\mathbb{Z}_2)$, as these give the same right ideal. This is just change of basis acting on the right by column operations; the result is called the theory of "elementary divisors", and you end up with the lemma. This is actually quite fun to do: if you like linear algebra, but you're used to doing it over a field, you might give it a try over $\mathbb{Z}_2$! In more generality, this is Exercise 26.6.
P.S. There is a typo in Example 17.6.3, it should map $\beta \mapsto \begin{pmatrix} 1-b_0 & 0 \\ 0 & b_0 \end{pmatrix}$ so that it has the right characteristic polynomial. But you get the same thing modulo $2$. (This is already in the errata.)