Let $X$ be a scheme of finite type over $\mathbb{Z}$. Let $R$ be the ring of algebraic integers. My intuition is that $X(R)$ is practically always infinite. 


More specifically, suppose that $X$ is faithfully flat over $\mathbb{Z}$, of relative dimension $\geq 1$, and the generic fiber is geometrically irreducible. Is that enough to guarantee infinitely many algebraic integer points?


This question is inspired by [this one](https://mathoverflow.net/questions/77604); I have no application in mind.



Are there nonobvious cases where equations have finitely many algebraic integer solutions?
This is basically true, in view of a density theorem due to Robert Rumely
(*Arithmetic over the ring of all algebraic integers*, J. reine u. angew. Math. **368**, 1986, p. 127-133). It relies on Rumely's capacity theory, and his extension of the theorem
of Fekete-Szegö.


For a generalization, and an algebraic proof, see also Laurent Moret-Bailly, *Groupes de Picard et problèmes de Skolem. II.* Annales scientifiques de l'École Normale Supérieure, Sér. 4, 22 no. 2 (1989), p. 181-194. (Numdam, <http://www.numdam.org/item?id=ASENS_1989_4_22_2_181_0>)


The hypothesis of Moret-Bailly's Theorem is that $X$ be irreducible, surjective and of positive relative dimension over ${\rm Spec}\mathbf Z$, and that its generic fiber be geometrically irreducible. Then, he proves that $X$ has $\overline{\mathbf Z}$-points which can be chosen arbitrarily close to a given $p$-adic point (end even more...).


