TITLE: Polynomial with many integer but no other rational solutions?
QUESTION [29 upvotes]: Is it true that if a (multivariate) polynomial has infinitely many integer solutions, then it also has a non-integer rational solution?

My motivation comes from this approach to Hilbert's tenth problem for the rationals.
There it was shown that the (non-polynomial) $2^{x^2}-y$ only has integer solutions over $\mathbb Q$.
I'm also interested to know what happens if we only want to force some of the variables to be integers, like just $x$ in $2^x-y$, but $y$ can take other rational values too.

REPLY [4 votes]: Expanding on some of the comments, I think that Vojta's conjecture will imply a strong result related to your question. First I'm going to reformulate the question in terms of Zariski density, which I think is more natural, since it can be applied inductively to reduce to the case of lower dimensional varieties.
Set-Up: Let $K$ be a number field, let $R$ be a ring of $S$-integers in $K$, let $X/K$ be a (smooth) projective variety, let $H\subset X$ be an ample divisor defined over $K$, and let $U=X\smallsetminus H$ be the affine subvariety of $X$ that is the complement of $H$. Choose an affine embedding $\phi:U\hookrightarrow \mathbb A^n$ defined over $K$. This allows us to talk about the $R$-integral points of $U$ relative to $\phi$, which we denote by
$$ U_\phi(R) = \phi^{-1}\bigl( \phi(U)(R) \bigr). $$
Question: Does
$$ \text{$U_\phi(R)$ is Zariski dense in $U$} \quad\Longrightarrow\quad
\text{$X(K)\smallsetminus U_\phi(R)$ is Zariski dense in $X$?} $$
Observation: Assume that Vojta's conjecture is true. If $X$ has Kodaira dimension $\kappa(X)\ge0$, then the question has an affirmative answer, for the silly reason that the conjecture implies in this case that $U_\phi(R)$ is never Zariski dense in $U$! More generally, again assuming Vojta's conjecture, the question has an affirmative answer for the same silly reason if there is an integer $m\ge1$ such that $\mathcal K_X\otimes\mathcal O_X(nH)$ is ample.
So if there are any examples for which the question has a negative answer, they must (conjecturally) be found using a variety of Kodaira dimension $-\infty$ and an effective ample divisor $H$ that isn't "too" ample.