To solve this problem, it is not necessary to compute powers of $I$ explicitly in terms of generators from knowing generators of $I$. Instead we can focus on the norm of powers of $I$ and reason indirectly by factoring ideals into prime ideals.
The ideal $I$ has prime norm $3$, so it is a prime ideal. (The converse is false: prime ideals need not have prime norm.) So $I^2$ has norm $9$ and $I^3$ has norm $27$.
Are there principal ideals with norm $3$? Let $\alpha = (1 + \sqrt{-23})/2$. For integers $m$ and $n$,
$$
{\rm N}(m + n\alpha) = {\rm N}\left(m + n\frac{1+\sqrt{-23}}{2}\right) = \frac{(2m+n)^2 + 23n^2}{4},
$$
so if ${\rm N}(m+n\alpha) = 3$ then $(2m+n)^2 + 23n^2 = 12$.
The left side is a sum of nonnegative terms, so the only choice is $n = 0$. Then $(2m)^2 = 12$, which has no integral solution.
Now we turn to $I^2$. It has norm $9$. Can $I^2$ be principal?
If $(m + n\alpha)$ has norm $9$ for some integers $m$ and $n$, then
$$
9 = {\rm N}(m + n\alpha) = \frac{(2m+n)^2 + 23n^2}{4},
$$
so $(2m+n)^2 + 23n^2 = 36$. This implies $|n| \leq 1$. If $n = \pm 1$ then $(2m+n)^2 = 36 - 23 = 13$, which is impossible. Thus $n = 0$, so $(2m)^2 = 36$, which tells us $m = \pm 3$. Thus $(m+n\alpha) = (\pm 3) = (3)$.
Can $I^2 = (3)$? No, because we can determine the prime ideal factorization of $(3)$ from the factorization mod $3$ of the minimal polynomial $T^2 - T + 6$ for $\alpha$: $T^2 - T + 6 \equiv T(T-1) \bmod 3$, which is a product of distinct monic irreducibles, so $(3) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ are distinct prime ideals. So $(3)$ is not the square of the prime ideal $I$ (unique factorization!) and either $\mathfrak p$ or $\mathfrak q$ is $I$.
Now we turn to $I^3$. It has norm $27$. If a principal ideal $(m+n\alpha)$ has norm $27$ then
$$
27 = {\rm N}(m + n\alpha) = \frac{(2m+n)^2 + 23n^2}{4},
$$
so $(2m+n)^2 + 23n^2 = 108$. That implies $|n| \leq 2$, and trying each option gives us the solutions
$$
(m,n) = (1,2), (-3, 2), (3,-2), (-1,-2),
$$
so $(m+n\alpha)$ is $(1+2\alpha) = (2+\sqrt{-23})$ or it is $(3-2\alpha) = (2-\sqrt{-23})$. We have shown the only principal ideals in $\mathcal O_K$ with norm $27$ are $(2\pm \sqrt{-23})$. Thus we want to factor these into prime ideals to see if either of them is $I^3$.
Since the ideals $(2+ \sqrt{-23})$ and $(2-\sqrt{-23})$ have norm $27$, their only prime ideal factors in $\mathcal O_K$ are prime ideal factors of $3$ (see my post here). We noted above that $(3) = \mathfrak p\mathfrak q$ for distinct prime ideals $\mathfrak p$ and $\mathfrak q$ (where $I$ is one of them). So the only possible ideals in $\mathcal O_K$ with norm $27$ are
$$
\mathfrak p^3, \ \ \mathfrak p^2\mathfrak q, \ \
\mathfrak p\mathfrak q^2, \ \ \mathfrak q^3.
$$
The ideal $(2+\sqrt{-23})$ is not divisible by both $\mathfrak p$ and $\mathfrak q$, since if it were then it would be divisible by their product $(3)$, and if $(3) \mid (2+\sqrt{-23})$ as principal ideals then $3 \mid (2+\sqrt{-23})$ as elements, but $2+\sqrt{-23}$ is not $3(a + b(1+\sqrt{-23})/2)$ for integers $a$ and $b$. Therefore the only possible
prime ideal factorization of $(2+\sqrt{-23})$ is $\mathfrak p^3$ or $\mathfrak q^3$. The exact same reasoning shows $(2-\sqrt{-23})$ is $\mathfrak p^3$ or $\mathfrak q^3$. And since the ideals $(2+\sqrt{-23})$ and $(2-\sqrt{-23})$ are different (they're principal ideals whose generators don't have a ratio that's a unit in $\mathcal O_K$, or in fact even lies in $\mathcal O_K$), their prime ideal factorizations are different. Thus if we let $\mathfrak p$ be the prime ideal such that $(2+\sqrt{-23}) = \mathfrak p^3$, then $(2-\sqrt{-23}) = \mathfrak q^3$.
Since $I$ is $\mathfrak p$ or $\mathfrak q$, we get $I^3 = (2+\sqrt{-23})$ or $I^3 = (2-\sqrt{-23})$. Which one is it?
Recall that $I = (3,1+\sqrt{-23})$, so easily
$$
2-\sqrt{-23} = 3 - (1+\sqrt{-23}) \in I.
$$
Thus $(2-\sqrt{-23}) \subset I$, so $I \mid (2-\sqrt{-23})$. Therefore $(2-\sqrt{-23}) = I^3$.
