The Discriminant of the $\mathbb{Q}$-basis $\{1, \alpha, \alpha^2\}$ is $-31$ which is square free, and contained in $\mathcal{O}_K$. From this can you conclude that $\{1, \alpha, \alpha^2\}$ is a basis for $\mathcal{O}_K$? You should be able to (this is a standard result).
If you convince yourself that the ring of integers is indeed $\mathbb{Z}[\alpha]$, then try to use the following (also well-known ?) result:
Let $p$ be a rational prime. Now $f = x^3 + x + 1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Reduce $f$ mod $p$ to obtain $\overline{f}$, and let $\overline{f} = \prod_i h^{e_i}_i$, be an irreducible factorization of $\overline{f}$ in $\mathbb{F}_p[x]$. Let $f_i \in \mathbb{Z}[x]$ such that $f_i$ mod p $= h_i$. Then the ideal $P_i = (p, f_i(\alpha))$ is prime in $\mathcal{O}_k$, and \begin{equation} (p) = \prod_i P^{e_i}_i \end{equation}
From your observation that $M_K \leq 1.57$, you further know that $\mathcal{O}_K$ is a PID, so all you $P_i$ are principal.