This question is only a slight variation from this question, but since rings and ideals are new to me I am struggling to apply Moos' argument to my current problem.
The problem is
If $K = \mathbb{Q}(\sqrt{7})$, then find all ideals $\mathfrak{a}$ of $\mathcal{O}_{K}$ such that $\mathcal{O}_{K}/\mathfrak{a}$ has order $6$. $\hspace{45pt}$($10$ marks)
I understand that this corresponds to finding the ideals of norm $6$ in $\mathcal{O}_{K}$. I first noticed that $$ 6 = (\sqrt{7} + 1)(\sqrt{7} - 1) $$ so that (using angle brackets to indicate the ideal generated by the contents) $$ \langle 6 \rangle = \langle\sqrt{7} + 1\rangle\langle\sqrt{7} - 1\rangle. $$ The embeddings of $K \hookrightarrow \mathbb{C}$ are determined by $\sqrt{7} \mapsto \sqrt{7}$ and $\sqrt{7} \mapsto -\sqrt{7}$, so that the norms of the above ideals are (noting that they're principal, so the norms of the ideals are equal to the absolute value of the norms of the generators) $$ \mathrm{Norm}(\sqrt{7} \pm 1) = |(\sqrt{7} \pm 1)(-\sqrt{7} \pm 1)| = 6. $$ Thus, both of $\langle\sqrt{7} + 1\rangle, \langle\sqrt{7} - 1\rangle$ are two such ideals.
To try and find some more, I then used to fact that $6 = 2 \times 3$ so that $$ \langle 6 \rangle = \langle 2 \rangle\langle 3 \rangle. $$ By the Kummer-Dedekind Theorem and noting that the minimum polynomial of $\sqrt{7}$ is $X^{2} - 7$, these ideals can be factorised into $$ \langle 2 \rangle = \langle 2, \sqrt{7} - 1 \rangle^{2} \hspace{20pt}\text{and}\hspace{20pt} \langle 3 \rangle = \langle 3, \sqrt{7} - 1 \rangle\langle 3, \sqrt{7} + 1 \rangle, $$ with $$ \mathrm{Norm}\big(\langle 2, \sqrt{7} - 1 \rangle\big) = 2 \hspace{20pt}\text{and}\hspace{20pt} \mathrm{Norm}\big(\langle 3, \sqrt{7} - 1 \rangle\big) = \mathrm{Norm}\big(\langle 3, \sqrt{7} + 1 \rangle\big) = 3 $$ also by K-D. Now we have $$ \langle 6 \rangle = \langle 2, \sqrt{7} - 1 \rangle^{2}\langle 3, \sqrt{7} - 1 \rangle\langle 3, \sqrt{7} + 1 \rangle. $$ I checked that $$ \langle 2, \sqrt{7} - 1 \rangle\langle 3, \sqrt{7} - 1 \rangle = \langle \sqrt{7} - 1 \rangle \hspace{20pt}\text{and}\hspace{20pt} \langle 2, \sqrt{7} - 1 \rangle\langle 3, \sqrt{7} + 1 \rangle = \langle \sqrt{7} + 1 \rangle, $$ so I didn't find any other ideals of norm $6$. Is this enough to show that $\langle\sqrt{7} + 1\rangle, \langle\sqrt{7} - 1\rangle$ are the only two ideals of $\mathcal{O}_{K}$ of norm $6$, by the unique factorisation of ideals?
It’s just a bit more complicated than @xarles suggests. But their advice to forget about ideals and work with elements instead is good, considering that your ring is principal.
The ramified primes are $2$ and $7$, with $2\sim(3-\sqrt7\,)^2$ and $7=\sqrt7\,^2$, where “$\sim$” means related by a unit multiplier. On the other hand, $3$ splits, with $3\sim(2-\sqrt7\,)(2+\sqrt7\,)$ As you can see, the (absolute values of the) respective norms are $2$ and $3$. To get something of norm $6$, you need to multiply something of norm $2$ to something of norm $3$. Only one choice for $2$, two choices for $3$, and you see that $(3-\sqrt7\,)(2-\sqrt7\,)\sim\sqrt7+1$, while $(3-\sqrt7\,)(2+\sqrt7\,)=\sqrt7-1$. The story would have been perhaps more interesting if both $2$ and $3$ had split.