Does a curve have infinitely many $K$-rational points under these hypotheses? The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?
Motivation
The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)
 A: Mazur in his article "Rational points of abelian varieties with values in towers of number fields", Invent. Math. 18 has produced lots of curves over number fields that have a finite number of points over $\mathbb{Z}_p$ extensions.
Also, there is a theorem due to Kato, Ribet, and Rohrlich that if $E/\mathbb{Q}$ is an elliptic curve and $K$ is the maximal abelian extension of $\mathbb{Q}$ unramified outside a fixed finite set of primes, then $E(K)$ is finitely generated. I imagine that if $E$ was instead hyperelliptic, then the conclusion would be that $E(K)$ is finite.
A trickier question is the following: a field $K$ is called ample if every smooth curve over $K$ that has a $K$-rational point has infinitely many of them (this is a stronger condition than what you want, since you only want this to be true for curves defined over some subfield of $K$). It seems that it is quite difficult to show that a concrete infinitely generated algebraic extension of $\mathbb{Q}$ is non-ample. For example it isn't even known whether $\mathbb{Q}^{ab}$ is ample. You might be interested in this survey on open problems concerning ample fields.
A: Let $X$ be the plane curve defined by the equation $x^2+y^2=0$. Let $k=\mathbb Q$ and $K=\mathbb R$.
The curve has exactly one point in both.
