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 $x,y,n=\pm 1$.

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

So our final system of equations is $(m-1)(m+1)=0,x^2-m n^2y^2=1, (x^2-1)(y^2-1)(n^2-1) z = 1$ 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.

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.