Field extensions over which algebraic varieties cannot acquire points

Question

The following fact (slightly reworded here) is proven in this answer:

If $K$ is a purely transcendental extension of an infinite¹ field $k$, then whenever a separated scheme $X$ of finite type over $k$ has points over $K$, it already has points over $k$ (viꝫ., $X(k) = \varnothing$ implies $X(K) = \varnothing$).

Let us say that the field extension $K$ of $k$ is “astigmagenic”² when the conclusion of this statement (the part that follows the “then”) holds. Thus, the above states that purely transcendental extensions of infinite fields are astigmagenic. But they are not the only ones: if $k$ is algebraically closed, then any extension of $k$ is astigmagenic (and it need not be purely transcendental: consider the field of functions of an algebraic variety that isn't rational).

Question: Can we characterize these “astigmagenic” field extensions in a simpler way? (E.g., do the finite type ones coincide with rational function fields of algebraic varieties for which the rational points are Zariski-dense?)

Note: I used “separated scheme of finite type” in the definition because this is what the linked answer does. But we can also try variations around this: e.g., maybe say that $K$ is weakly astigmagenic when $X(k) = \varnothing$ implies $X(K) = \varnothing$ for $X$ a geometrically integral separated scheme of finite type. It is very unclear to me how much this changes the condition, so I'm interested in the relations between the various variations: feel free to answer with whatever variation seems to make the question most natural or interesting.

1. As pointed out by Laurent Moret-Bailly in a comment, the word “infinite” was missing from the answer.

2. Pardon my Greek. This is supposed to mean “that does not create points”.

Answer 1

Let $K$ be a finite type field extension of $k$ which corresponds to a rational function field of an algebraic variety $V$ for which the rational points are not Zariski dense.

Let $U$ be the complement in $V$ of the Zariski closure of the rational points.

Then $U$ has no $k$-point, but has a $K$-point (the generic point), so $K/k$ is not astigmagenic.

Since every finite type extension is the rational function field of a variety, it follows that a finite type extension is astigmagenic if and only if it is the rational function field of an algebraic variety for which the rational points are Zariski-dense. (Well, that proves "only if", and "if" follows from the same proof as for purely transcendental - given a $K$-point, we obtain a rational map from the variety, hence from an open subset of the variety, and then look at the image of a rational point).