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.

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