Are there infinite prime divisors of sequence $a_{n}=2^{F_{0}}+2^{F_{1}}+\dots 2^{F_{n}}$ We define the sequence of Fibonacci numbers $(F_n)_{n \geqslant 0}$ as follows:
$$F_n=
\begin{cases}
0 & \text{if $n=0$} \\
1 & \text{if $n=1$} \\
F_{n-1}+F_{n-2}& \text{if $n>1$}
\end{cases}$$
The sequence goes as follows : $0,1,1,2,3,5,8,13,\ldots$
Furthermore, we define $(a_n)_{n \geqslant 0}$ as follows:
$$a_n=\sum_{i=0}^n 2^{F_i} = 2^{F_0}+2^{F_1}+\ldots+2^{F_n}$$
The sequence goes as follows : $1,3,5,9,17,49,305,8497,\ldots$

Question : Are there infinitely many primes dividing some element of $(a_n)$?


Haran's edit:
I believe that this might be related to Kobayashi's Theorem:

Let $M$ be an infinite set of positive integers such that the set of prime divisors of the numbers in $M$ is finite. Then, the set of primes dividing the numbers in the set:
$$M+a=\{m+a \mid m \in M \}$$
is infinite, where $a$ is a non-zero fixed integer.

 A: Great question! First, let a prime $p \mid a_n$. We can see that $a_0=2^{F_0}=2^0=1$ and thus $a_0$ is odd. Furthermore, $2^{F_n}$ is even for $n \geqslant 1$, and thus $a_n$ is odd for all $n \in \mathbb{Z_0}$ (non-negative integers). This means that $p$ is an odd prime.

Now, observe the Fibonacci numbers modulo $m$. We have $F_0 \equiv 0 \pmod{m}$, $F_1 \equiv 1 \pmod{m}$ and by the recursive formula for Fibonacci Numbers:
$$F_{n+2} \equiv F_{n+1}+F_n \pmod{m}$$
where $n \in \mathbb{Z_0}$. It is easy to see by the Pigeonhole Principle that $(F_{n+1},F_n) \bmod{m}$ must repeat. Then, we can retrace using $F_{n-1} \equiv F_{n+1}-F_n \pmod{m}$ to see that we have $F_x \equiv 0 \pmod{m}$ and $F_{x+1} \equiv 1 \pmod{m}$ for some $x \in \mathbb{N}$. The smallest such $x$ becomes the period of $F \bmod m$ and is known as the Pisano Period. We shall make use of the same in our proof.

It is clear that $2^{p-1} \equiv 1 \pmod{p}$ by Fermat's Little Theorem. Thus, we have:
$$x \equiv y \pmod{p-1} \implies 2^x \equiv 2^y \pmod{p}$$
Now, let $\pi(p-1)$ be the Pisano Period of the Fibonacci Numbers modulo $p-1$. Consider $a_k$ for $p\pi(p-1) \mid k$. We have:
$$a_k=\sum_{i=0}^k 2^{F_i} = 2^{F_k}+\sum_{i=0}^{k-1} 2^{F_i} = 2^{F_k}+\frac{k}{\pi(p-1)}\sum_{i=0}^{\pi(p-1)-1} 2^{F_i} \equiv 2^{F_k} \equiv 1\pmod{p}$$
as $p \mid \frac{k}{\pi(p-1)}$ and $F_k \equiv 0 \pmod{p-1}$ since $\pi(p-1) \mid k$. Thus, we have:
$$p\pi(p-1) \mid k \implies a_k \equiv 1 \pmod{p}$$

Now, assume that there are only finitely many possible prime divisors for $(a_n)$. Let the set of all such prime divisors be $S$. Simply consider:
$$k=\prod_{p \in S} p\pi(p-1) \implies a_k \equiv 1 \pmod{p}$$
for every $p \in S$. This is clearly a contradiction as $a_k$ does not have any prime factor since none of the primes in $S$ divide $a_k$.
Hence, the sequence $(a_n)$ must contain infinitely many prime factors, and we are done!
