Class number of $\mathbb{Q}[x]/(x^{3} - 3x+1)$ is 1

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This is an exercise in Milne's algebraic number theory. Let $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of $x^{3}- 3x+1$. By following hints, I showed that $\mathcal{O}_{K} = \mathbb{Z}[\alpha]$ and also I know that $(2)$ inerts in $\mathcal{O}_{K}$. The author said that one can show $K$ has class number 1 (i.e. PID) from this, but I can't figure out how to proceed. The Minkowski's bound is 40.5, so I can try to figure out how the primes less than 40 behave, but I don't think this is what Milne wants to do. Any ideas?

To show $\mathcal{O}_{K} = \mathbb{Z}[\alpha]$, we showed that $3\mathcal{O}_{K} = (\alpha+1)^{3}$, so $(\alpha + 1)$ is a prime ideal with $\mathcal{O}_{K} = \mathbb{Z}[\alpha] + (\alpha + 1)\mathcal{O}_{K}$. Since the discriminant of $\mathbb{Z}[\alpha]$ is $81$, we have $3^{4} \mathcal{O}_{K} \subseteq \mathbb{Z}[\alpha]$, and this proves the claim.