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"][1], and Lemma 9 in [Tao's blog][2].

**Added.** By (8) in the [quoted article][1], 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.

  [1]: http://retro.seals.ch/cntmng?type=pdf&rid=ensmat-001:1974:20::62&subp=hires
  [2]: https://terrytao.wordpress.com/2014/05/13/dworks-proof-of-rationality-of-the-zeta-function-over-finite-fields/