Context-free grammar for $L = \{a^{2^k}, k \in\mathbb{N}\}$

In an exercise, I am asked to find a context free grammar for language $L = \{a^{2^k}, k \in \mathbb{N}\}$.

I have been trying to use a "doubling" variable. If $a^{2n} \in L, n\in\mathbb{N}$ then use this variable to double the $a$'s that have been produced by the other language rules.

Is this thinking valid? So far I haven't been able to get anywhere with it, but it seems logical given the double-stack of powers.

$L = \{a^{2^k}, k \in \mathbb{N}\}$ is not a context-free language according to Pumping lemma for context-free languages.

Suppose $L$ is context-free. The pumping lemma says there exists some integer $p \ge 1$ such that every string $s$ in $L$ where $|s| \ge p$ can be written as $s=uvwxy$ where $|vwx|\le p$, $|vx|\ge 1$ and $uv^nwx^ny$ is in $L$ for all $n \ge 0$.

Let $s$ be a string in $L$ longer than $p$, and $u$, $v$, $w$, $x$, and $y$ have the properties given by the pumping lemma. Thus $uwy, uvwxy, uv^2wx^2y\in L$. Let $a$ and $b$ be such that $$|uwy|=2^a, |uvwxy|=2^b$$ Note $b>a$. Then $$|uv^2wx^2y|=2|uvwxy|-|uwy| = 2^{b+1}-2^a = 2^a(2^{b+1-a}-1)$$ But $2^a(2^{b+1-a}-1)$ is not a power of 2, and so $uv^2wx^2y\notin L$.

StackExchange Computer Science Q#47137, answer score: 11