Yes. Consider the set $U$ of pairs $(x,e^x)$ where $x \in \mathbb{Z}.$ Then no$x \in \mathbb{Z}$ and let $U' $ be an infinite subset of $U$$U.$ Then $U'$ is not contained in the vanishing of aany nonzero polynomial in $\mathbb{R}[X,Y].$ As giventhe vanishing of any polynomial $F\in\mathbb{C}[X,Y],$$F\in\mathbb{C}[X,Y]$ is contained in the productvanishing of $F$ and its coordinatewise congugate $\overline{F}$ is asome polynomial overin $\mathbb{R}$ containing$\mathbb{R}[X,Y]$, namely the vanishingproduct of F;F and its coordinatewise conjugate, it follows that no infinite subset of that $U$$U'$ is thenot contained in the vanishing of aany nonzero polynomial in $\mathbb{C}[X,Y].$ Hence
Consider the closure of $U' $ denoted by $\overline{U'} $. As a Zariski closed set, if $U'$$\overline{U'} $ is an infinite subsetthe union of $U$ and $\overline{U'}$ is its closure, we may choosefinitely many varieties. Choose $V$ to be an irreducible component of $\overline{U'}$ containing infinitely many points of $U'.$ The variety $V$ is infinite and therefore of dimension greater than 0. Furthermore, by our above remarks, $V$ is not the vanishing of any nonzero polynomial over $\mathbb{C}.$ As every prime ideal in $\mathbb{C}[X,Y]$ of codimension 1 is principal, it follows that $dim V = 2$$dim V \neq 1.$ We conclude $dimV = 2$ and $\mathbb{A}^2 = V = \overline{U'}.$
From this we obtain that $\overline{U'} = \mathbb{A}^2.$$U$ is dense in $\mathbb{A}^2$ and the closure of any subset of U is either finite or all of $\mathbb{A}^2,$ as desired.