Per Gazerun's comment answer to related question which leads 
to relaxation of the OP.

This is possible over the integers for all $d$ 
and the question can be relaxed
by allowing finitely many points over the rationals.

First we define the set $\{1,-1\}$ with the equation $(m-1)(m+1)=0$.

The only integers solutions to $x^2-m n^2y^2=1$ are 
$x=\pm 1, ny=0$ and $y=\pm 1,x=0$ and $n=\pm 1,x=0$.

Over $\mathbb{Z}[\sqrt{d}]$ for $n=\sqrt{d}$ this is 
Pell equation $x^2 \pm ny^2=1$ with infinitely many solutions and if $d$
is square we are in the above case.

So we must get rid of the bad points $x=\pm 1,0$.

Use the following cheap trick: $ (x^2-1)x z = 1$. This is linear in $z$
unless $(x^2-1)x=0$ which leads to $0=1$.

So our final system of equations is 
$(m-1)(m+1)=0,x^2-m n^2y^2=1, (x^2-1)x z = 1$ which doesn't have integer solutions
but have infinitely many over $\mathbb{Z}[\sqrt{d}]$ for all 
$d$. If $d$ is negative chose $m=-1$ otherwise $m=1$

---

Partial result, some one could try to extend it.

It is possible to define variety the complement of $x=y^2$,
that is $x \ne y^2$:

$$ zt=1 \qquad (1)$$
$$ (x-y^2) z=1 \qquad (2)$$.

In (1) and (2) $z$ can be any nonzero rational. $x\ne y^2$ for
obvious reasons and $x-y^2=1/z$.

Appears to me this allows to describe the complement of variety 
given by single equation.