The function $$ f(z)=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} $$ defines a holomorphic function for $|z|<2$, and it satisfies $$ f(2z) = f(z)+\frac{z}{1-z} $$ for $|z|<1$. Based on this identity, it is easy to prove that $f(z)$ extends to a meromorphic function on $\mathbb{C}$, and the set of poles is $\{2^n:\ n=1,2,\dots\}$. In particular, $f(z)$ does not define a rational function, because its meromorphic extension to $\mathbb{C}$ has infinitely many poles.
Regarding your second question, I recommend the work of Dwork (with which I am not familiar), e.g. (8) in Alain Robert's article "Des adèles: pourquoi", and Lemma 9 in Tao's blog.
Added 1. By (8) in the quoted article, one can give a number theoretic proof: the Taylor coefficients of $f(z)$ around the origin are rational, but their denominators $2^n-1$ are not supported on finitely many primes. (Indeed, if $p$ is a prime, then every prime factor of $2^p-1$ is greater than $p$.) This proof relies on the work of Fatou only, the subsequent work of Dwork is not needed.
Added 2. A more direct proof of my claims above follows from the formula $$ \sum_{m=1}^\infty\frac{z}{2^m-z} = \sum_{n=1}^\infty \frac{z^n}{2^n-1},\qquad |z|<2. $$ Indeed, left hand side defines a meromorphic function on $\mathbb{C}$ with pole set $\{2^m:\ m=1,2,\dots\}$.