Let $k$ be a perfect field, and let $\bar k$ be a fixed algebraic closure of $k$.
Let $\overline{X}$ be a nonempty smooth algebraic variety over $\bar k$.
Does there exist a natural number $d=d(\overline{X})$ with the following property:



> 
> For any $k$-form $X$ of $\overline{X}$, the variety $X$ has a $K$-point over some finite field extension $K$ of $k$ of degree $[K:k]\le d$ ?
> 
> 
> 


The answer YES would imply Theorem 2 of my answer to [this question](https://mathoverflow.net/questions/239022/).


[This question](https://mathoverflow.net/questions/128388/) and the references in comments to it may be relevant.



Algebraic points of uniformly bounded degree on an algebraic variety
No: this fails already when $X=E$ is an elliptic curve and $k=\mathbb{Q}$. This would imply that every element of $H^1(\mathbb{Q},E)$ has order at most $d$, and I'm pretty sure that this cohomology group has elements of arbitrarily large order. Otherwise the Tate-Shafarevich conjecture would be rather trivial, as it is well-known that $Sha(E)[d]$ is finite for every $d$.


