Polynomial with many integer but no other rational solutions? 
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.
 A: 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.
A: This is a partial answer which highlights some of the subtleties of this question. I will use algebraic geometry language as this is the correct set up for such questions.
First, this question is only really interesting for affine varieties as for projective varieties, every rational point is an integral point.
I take the following set-up. Let $U$ be a smooth affine variety over $\mathbb{Q}$. I consider the opposite question: does there exist $U$ such that $U(\mathbb{Z})$ is infinite and $U(\mathbb{Z}) = U(\mathbb{Q})$?
So I show there no such examples in dimension $1$ and discuss dimension $2$, assuming some standard conjectures. I let $U \subset X$ be a smooth projective compactification.
First recall that we say that $U$ satisfies weak approximation at a prime $p$ if $U(\mathbb{Q})$ is dense in $U(\mathbb{Q}_p).$
Lemma
If $U$ satisfies weak approximation at some large enough prime $p$, then $U(\mathbb{Z}) \neq U(\mathbb{Q})$.
Proof
By the Lang-Weil estimates, for all sufficiently large primes $p$ we have $U(\mathbb{F}_p) \neq X(\mathbb{F}_p)$. Providing $U$ satisfies weak approximation at $p$, I can therefore choose a rational point whose reduction modulo $p$ lies in $(X \setminus U)(\mathbb{F}_p)$. Such a rational point does not even lie in $U(\mathbb{Z}_p)$, let alone $U(\mathbb{Z})$.
Now consider the case $\dim U = 1$. If $g(X) = 0$ then $U$ satisfies weak approximation at all primes. If $g(X) \geq 2$, then $X(\mathbb{Q})$ is finite by Faltings' theorem. If $g(X) = 1$ then $U(\mathbb{Z})$ is finite by Siegel's Theorem. So here there is no example.
Now consider the case $\dim U = 2$. I expect this question is a birational invariant, so we may assume that $X$ is minimal, so we just go through the Enriques–Kodaira classification of surfaces. I won't do this in detail here, but it seems like no examples can arise.
For example, any geometrically rational or K3 surface is conjectured to satisfy weak approximation at all sufficiently large primes, so these arn't allowed. (This rules out a potential suggestion from the comments). If $X$ has general type, then conjecturally $X(\mathbb{Q})$ is not Zariski dense. We thus reduce to the case of dimension $1$, which was covered above.
A: This is just a long comment, but it feels natural to look
at the $a_{ij}$ defined via
$$
\sum_{i,j \geq 0} a_{i,j} x^i y^k = \frac{1}{P(x,y)}
$$
where $P(x,y)=0$ has infinitely many integer solutions.
(Assuming (0,0) is not a solution).
