As long as you allow a fixed number field $F = {\bf Q}(\alpha)$ you can prove $H^{o(1)}$ as you more-or-less suggest towards the end, by first showing that the number of ideals of $F$ that divide $n$ is $H^{o(1)}$ and then proving that any ideal has $O(\log^r H)$ generators of height at most $H$, where $r = r_1 + r_2 - 1$ is the rank of the unit group $U_F$ of $F$.
The first part is basically the same as the argument over $\bf Z$. If the ideal $(n)$ factors into prime powers as $\prod_i \wp_i^{e_i}$ then there are $\prod_i (e_i + 1)$ factors. ideals that divide $n$. Given $\epsilon > 0$ there are finitely many choices of rational prime $p$ and integer $e>0$ such that $e+1 > (p^e)^\epsilon$, and therefore only finitely many choices of a prime $\wp$ of $F$ and $e>0$ such that $e+1$ exceeds the $\epsilon$ power of the norm of $\wp^e$. Therefore $\prod_i \wp_i^{e_i} \ll_\epsilon N^\epsilon$ where $N$ is the absolute value of the norm of $n$. But $N \ll H^{[F : {\bf Q}]}$, and $\epsilon$ was arbitrary, so we've proved the $H^{o(1)}$ bound.
For the second part, Dirichlet gives a logarithm map $U_F \rightarrow {\bf R}^r$ whose kernel is finite (the roots of unity in $F$) and whose image is a lattice $L$. A unit of height at most $H$ has all its conjugates of size $O(H)$, so is mapped to a ball of radius $\log(H) + O(1)$. Therefore there are $O(\log^r(H))$ units of height at most $H$. Much the same argument (involving a translate of $L$) shows that he same bound applies also to generators of any ideal $I$, since the ratio of any two generators of $I$ is a unit.
[I see that "Pitt the Elder" just gave much the same answer.]
