Variety acquiring rational point over any quadratic extension Does there exist a variety $X$ over $\mathbb{Q}$ (or a number field) such that it has no rational points over $\mathbb{Q}$ but acquires points over any quadratic extension $\mathbb{Q}(\sqrt{d})$?
If there is an example, how often can this happen?
Can we generalize to an extension of fixed degree?
(This might be a stupid question and I'm guessing answer is no, but I couldn't show it)
 A: According to this answer by Laurent Moret-Bailly, the set of rational non-squares is diophantine over the rationals (a result of Bjorn Poonen): 
there is a polynomial $P(a,x_1,\dots,x_n)$ which for $a\in\mathbb{Q}$ has a rational solution
iff $a$ is a rational non-square. For the variety, take the hypersurface with equation $P^2+(x_{n+1}^2-a)^2=0$ in $n+2$ coordinates $(a,x_1,\dots,x_{n+1})$.
A: 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.
A: Consider a hyperelliptic curve $y^2=f(x)$. I think by Falting's theorem on subvarieties of abelian varieties, we know that all but finitely many points of this over quadratic subfields have $x \in \mathbb Q$.
So this curve is such a variety if and only if the equation $y^2 D = f(x)$ has solutions for all but finitely many squarefree $D$.
Heuristically this shouldn't happen as the squarefree part of a random number $n$ is rarely much smaller than $n$, so the squarefree parts of $f(x)$ should grow quickly with $x$, because $f(x)$ has degree at least $3$, and so most numbers should not appear. In other words, because most hyperelliptic curves have no rational points, we expect the family $y^2 D = f(x)$ to mostly have no rational points, unless there is something strange going on with $f$.
But can this property actually be established for hyperelliptic curves?
A: Will Sawin and Michael Stoll have noted that, as a consequence of Faltings's "Big Theorem," a hyperelliptic equation $y^2 = f(x)$ with $\deg{f} > 6$ (genus $> 2$) and not admitting a degree $2$ non-constant map to an elliptic curve, has all but finitely many of its quadratic solutions $x, y \in \mathbb{Q}(\sqrt{d}), \, d \in \mathbb{Z}$, satisfy $x \in \mathbb{Q}$. We may add to this an argument due to Granville in Rational and integral points of quadratic twists of a given hyperelliptic curve to show:
Claim. The ABC conjecture implies, for a fixed $f$ having $\deg{f} > 6$ and no repeated roots, that the number of squarefree $d$ in $|d| \leq D$ for which the equation $dy^2 = f(x)$ has a rational solution with $y \neq 0$, is $O(D^{2/3+o(1)})$.
This answers Will's question under ABC. Hence ABC takes care of the problem for most $y^2 = f(x)$ - save for the ones of genus one or two or those doubly covering an elliptic curve. (For rational curves $C/\mathbb{Q}$ the problem is easy: Hasse's theorem shows that $C(\mathbb{Q}) = \emptyset$ is only possible when $C(\mathbb{Q}_p) = \emptyset$ for some prime $p$, but then $C$ will not have points in any quadratic field $\mathbb{Q}(\sqrt{d})$ split by $p$.) It seems to me that the case of hyperelliptic curves of genus $> 2$ doubly covering an elliptic curve can be settled under ABC by similar methods (Faltings's theorem and a modification of Granville's argument), whereas the genus one case should be solved unconditionally by Kolyvagin's theorem and non-vanishing results, cf. Chris Wurthrich's comment in the linked question. I am not sure about the genus two case though - it is barely missed by the argument below.
Proof of the claim. (Granville). M. Langevin has noted (cf. Thm. 12.2.12 in Heights in Diophantine Geometry by Bombieri and Gubler) that Elkies's construction for "ABC $\Rightarrow$ Roth" via Belyi maps yields in fact much more than Roth's theorem:
Lemma. Let $\varepsilon > 0$ and let $F \in \mathbb{Z}[X,Y]$ be a homogeneous polynomial with distinct linear factors over $\mathbb{C}$. Then for all co-prime $m,n$ with $F(m,n) \neq 0$, the (strong) ABC conjecture implies $\mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} \max(|m|,|n|)^{\deg{F}-2-\varepsilon}$.
The ABC conjecture is recovered as the special case $F(X,Y) = XY(X+Y)$, whereas Roth's theorem is just the weakening of this statement dropping the radical. 
We apply this is follows. Consider the $\asymp T^2$ rational values $x = m/n \in \mathbb{Q}$ with $T \leq \max(|m|, |n|) < 2T$, $(m,n)=1$, $n > 0$, and $f(x) \neq 0$. Writing $du^2 = F(m,n)$ with $F$ the homogenization of $f$, the above quoted form of ABC yields
$$
(DT^{\deg{f}})^{1/2} \gg_f |dF(m,n)|^{1/2} = |du| \geq \mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} T^{\deg{f} - 2 - \varepsilon},
$$
or $D > T^{\deg{f} - 4 - o(1)}$, in $|d| \leq D$. As the irreducible fraction $x$ uniquely determines the squarefree part $d$, we get what we want by splitting the range $T < D^{1/(\deg{f}-4-o(1))}$ into dyadic intervals.
