Is $\lim\_{n \rightarrow \infty} |\{(x,y) \in \mathbb{Q}(\zeta\_n)^2 : y^3 = x^3 + x + 1\}| < \infty ?$ where $\zeta\_n$ is a primitive $n$-th root of unity.


That is, I am asking whether the number of $\mathbb{Q}^{\text{ab}}$-rational points of $y^3 - x^3 - x - 1$ is finite.


If the answer is negative, then a slight augmentation of the polynomial to a one that makes the limit finite or a bound on the rate of growth will be very much appreciated. 



The number of solutions of a Diophantine equation
To answer your concrete question, first, your curve (or rather, its projective closure) is isomorphic to the elliptic curve 8649b1, $y^2 + y = x^3 - 8$. Already over $\mathbb Q$, it has rank two and therefore infinitely many rational points.


More generally, for any elliptic curve $E$ over $\mathbb Q$, you should be able to find a quadratic twist $E^{(d)}$ with positive rank; then $E({\mathbb Q}(\sqrt{d}))$ is infinite, and since ${\mathbb Q}(\sqrt{d})$ is contained in a cyclotomic field, your limit is infinite.


For curves of higher genus, the question might be more interesting.


If you have a *hyperelliptic* curve $C \colon y^2 = f(x)$, then you will get infinitely many points with rational $x$-coordinate and $y$-coordinate in a quadratic number field. Since the compositum of all quadratic fields is contained in ${\mathbb Q}^{\text{ab}}$, this shows that hyperelliptic curves also have infinitely many points over this field.


