You say "the ideal $I$ seems very odd", but in fact it is very even, since its norm is a multiple of $2$. :)
Anyway, the norm of an ideal is a very important numerical value for figuring out how an ideal factor factors.
Theorem. If $I$ is a nonzero ideal in $\mathcal O_K$ and its norm is divisible by a prime $p$, then some prime ideal in $\mathcal O_K$ dividing $p$ is a factor of $I$.
Proof. The ring $\mathcal O_K/I$ has size equal to the norm of $I$ (do you know that important combinatorial interpretation of the ideal norm?), so we are told $\mathcal O_K/I$ has size divisible by $p$. Viewing $\mathcal O_K/I$ merely as an additive group, since its order is divisible by $p$, Cauchy's theorem from group theory tells us that there is an $\alpha \bmod I$ with additive order $p$, so
$$
\alpha \not\equiv 0 \bmod I, \ \ \ p\alpha \equiv 0 \bmod I.
$$
That is equivalent to $I \nmid (\alpha)$ and $I \mid (p\alpha)$. Since $(p\alpha) = (p)(\alpha)$,
$$
I \nmid (\alpha), \ \ \ I \mid (p)(\alpha).
$$
If $I$ is not divisible by any prime ideal dividing $(p)$, then $I$ and $(p)$ are relatively prime ideals, so from $I \mid (p)(\alpha)$ we get $I \mid (\alpha)$, but that contradicts the property $I \nmid (\alpha)$. Thus $I$ must be divisible by some prime ideal factor of $(p)$. QED
Thus for your ideal, with norm $2 \cdot 3 \cdot 7 \cdot 23$, it must factor as a product of prime ideals with norms $2, 3, 7$, and $23$. A prime ideal has prime-power norm, so when a prime factor of the norm has multiplicity $1$, there must be a single prime ideal factor with that prime norm. Things would be more subtle if an ideal has norm divisible by a higher power of a prime, say $3^2$. Then the ideal might be divisible by two prime ideals of norm $3$ or a single prime of norm $9$.
Example. In $\mathbf Z[\sqrt{-5}]$, let $I = (7+\sqrt{-5})$. Then $I$ has norm $49 + 5 = 54 = 2 \cdot 3^3$, so the prime ideal factors of $I$ are among the prime ideal factors of $2$ and $3$.
In $\mathbf Z[\sqrt{-5}]$, $(2) = \mathfrak p^2$ where $\mathfrak = (2,1+\sqrt{-5})$ and $(3) = \mathfrak q\mathfrak q'$, where $\mathfrak q = (3,1+\sqrt{-5})$ and $\mathfrak q' = (3,1-\sqrt{-5})$. So $I$ must be divisible by $\mathfrak p$ (the only ideal of norm $2$). It can't be divisible by both $\mathfrak q$ and $\mathfrak q'$, by contradction: if $(7+\sqrt{-5})$ were divisible by $\mathfrak q$ and $\mathfrak q'$ then it would be divisible by their product $(3)$, and having $(3) \mid (7+\sqrt{-5})$ as principal ideals in $\mathbf Z[\sqrt{-5}]$ forces $3 \mid (7+\sqrt{-5})$ as elements in $\mathbf Z[\sqrt{-5}]$, but $7 + \sqrt{-5} \not= 3(m+n\sqrt{-5})$ for integers $m$ and $n$ since $7$ is not a multiple of $3$ (and $1$ isn't either). Thus the only way to explain $I$ having norm divisible by $3^3$ is to have $I$ be divisible by $\mathfrak q^3$ or $\mathfrak q'^3$, so
$$
I = \mathfrak p\mathfrak q^3 \ \ {\sf or } \ \
I = \mathfrak p\mathfrak q'^3.
$$
Which one is it? We just need to figure out if $\mathfrak q \mid I$ or $\mathfrak q' \mid I$, or equivalently,
$$
7+\sqrt{-5} \stackrel{?}{\equiv} 0 \bmod \mathfrak q \ \ {\sf or } \ \ 7+\sqrt{-5} \stackrel{?}{\equiv} 0 \bmod \mathfrak q'.
$$
Since $\mathfrak q = (3,1+\sqrt{-5})$, it's quite easy to write $7+\sqrt{-5}$ in terms of the generators of $\mathfrak q$:
$7 + \sqrt{-5} = 3 \cdot 2 + 1 + \sqrt{-5} \in \mathfrak q$, so $\mathfrak q \mid (7+\sqrt{-5})$. Thus
$$
(7+\sqrt{-5}) = \mathfrak p\mathfrak q^3 = (2,1+\sqrt{-5})(3,1+\sqrt{-5})^3.
$$
