Let $n > 1$ be odd.
The curve:

$$X:  \ x^n + 2  y^n + 4 z^n = 0$$

does not have any points over $K/\mathbf{Q}_2$ unless the ramification index $e(K/\mathbf{Q}_2)$ is divisible by $n$. <b>Proof:</b> At least two of the terms $x^n$, $2 y^n$, $4 z^n$ must have the same $2$-adic valuation. On the other hand, the ramification index $e$ of any abelian extension $K/\mathbf{Q}_2$ is a power of two. So $X$ has no points over any Galois extension of $\mathbf{Q}$ whose decomposition group at any prime above $2$ is abelian. In particular, it has no point over any cyclotomic extension. It has genus $>1$ if $n>3$.

You can ask whether any smooth projective curve $X/\mathbf{Q}$ has at least one rational point over <i>any</i> solvable extension. Since local Galois groups are solvable, there are no longer any local obstructions. This is an open problem, and a positive answer would have various nice consequences including (generalizations of) Serre's conjecture, etc. There's no particular reason why it should be true, however.