If $p | f(n)$ precisely, then one of the ideals $(n - \sigma(\alpha))\mathcal{O}_K$ must have a factor of $p\mathcal{O}_K$ in its factorization and hence all of them do. Therefore $p\mathcal{O}_K$ splits into a product of $deg(f)$ ideals and therefore they must all have degree one.

One could see quite elementarily that there exist infinitely many primes that divide some value of $f$ precisely: if $p$ is not such a prime, then if $p^2 | f(n)$, then $p^2 | f(n + p) = f(n) + pf'(n) + p^2A$ for some integer $A$ and therefore $p | f'(n)$. Thus $n$ is a common zero of both $f, f'$ in $\mathbb{F}_p$, which cannot happen for $p$ large enough, as $Res(f, f')$ does not vanish.

If $p \mid f(n)$ precisely, then one of the ideals $(n - \sigma(\alpha))\mathcal{O}_K$ must have a factor of $p\mathcal{O}_K$ in its factorization and hence all of them do. Therefore $p\mathcal{O}_K$ splits into a product of $\deg(f)$ ideals and therefore they must all have degree one.

One could see quite elementarily that there exist infinitely many primes that divide some value of $f$ precisely: if $p$ is not such a prime, then if $p^2 \mid f(n)$, then $p^2 \mid f(n + p) = f(n) + pf'(n) + p^2A$ for some integer $A$ and therefore $p \mid f'(n)$. Thus, $n$ is a common zero of both $f, f'$ in $\mathbb{F}_p$, which cannot happen for $p$ large enough, as $\mathrm{Res}(f, f')$ does not vanish.