Is this a rational function?

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. See also Remark 2 below.

Remark 1. A more direct proof of the above claims follows from the identity $$ \sum_{m=1}^\infty\frac{z}{2^m-z} = \sum_{n=1}^\infty \frac{z^n}{2^n-1},\qquad |z|<2. $$ Indeed, left hand side defines a meromorphic function on $\mathbb{C}$ with pole set $\{2^m:\ m=1,2,\dots\}$.

Remark 2. One can give a different, number theoretic proof using Eisenstein's theorem on algebraic functions (the proof was published by Heine because of Eisenstein's early death). Indeed, the Taylor coefficients of $f(z)$ around the origin are rational, but their denominators $2^n-1$ are not supported on finitely many primes by Fermat's little theorem. (As Gerald Edgar remarked below, this argument proves that $f(z)$ is not even algebraic.)

Let $F(z)=\sum_{n\geq 1}\frac{z^n}{2^n-1}$. Then $F(2z)-F(z)= \frac{z}{1-z}$. Assume that $F(z)$ is rational. Thus $F(z)$ has a
pole at some $z_0\neq 0$, and $F(2z)$ has a pole at $z_0/2$. Let $z_1$ be a pole of $F(z)$ of maximum absolute value. Let $z_2$ be a pole of $F(2z)$ of minimum absolute value. Then $z_2\neq z_1$, and both $z_1$ and $z_2$ are poles of $F(2z)-F(z)$, a contradiction.

The "standard" necessary and sufficient condition for a power series $\sum a_n z^n$ to be a rational function is that the infinite Hankel matrix $[a_{i+j}]_{i,j\geq 0}$ has finite rank, but I don't know if this can be applied to the present question. See Enumerative Combinatorics, vol. 1, 2nd ed., Exercise 4.6.

$\sum a_n z^n$ is a rational function iff $a_n$ is a sum of polynomials times exponentials. This is a straightforward corollary of partial fraction decomposition. So, suppose $\frac{1}{2^n - 1}$ can be expressed as such a sum. Taking $n \to \infty$ shows that the largest $r$, in absolute value, such that $r^n$ appears in this sum is $r = \frac{1}{2}$, and moreover (after multiplying both sides by $2^n$) that its polynomial coefficient must be the constant polynomial $1$. That is, the sum must begin

$$\frac{1}{2^n - 1} = \frac{1}{2^n} + \text{smaller terms}.$$

The next largest $r$ such that $r^n$ can apppear in this sum is determined by the asymptotic behavior of $\frac{1}{2^n - 1} - \frac{1}{2^n} = \frac{1}{2^n(2^n - 1)} \approx \frac{1}{4^n}$, and the same $n \to \infty$ and multiplying by $4^n$ argument as above shows that it must be $r = \frac{1}{4}$ with polynomial coefficient $1$. So the sum must continue

$$\frac{1}{2^n - 1} = \frac{1}{2^n} + \frac{1}{4^n} + \text{smaller terms}.$$

But it's clear that in fact we have

$$\frac{1}{2^n - 1} = \sum_{k \ge 1} \frac{1}{2^{kn}}$$

so this argument never terminates, and it follows that $\frac{1}{2^n - 1}$ cannot be expressed as a finite sum of polynomials times exponentials. This is a less complex-analytic version of the argument that proceeds by showing that the generating function has infinitely many poles.