Fibonacci numbers that are powers of 2

Fibonacci numbers have just about the greatest divisibility rule you could expect. Fibonacci numbers share common divisors exactly when their corresponding indices share common divisors, $\gcd(F[m],F[n])$ = $F_{\gcd(m,n)}$.

This result means that the Fibonacci index of any power of $2$ greater than $8$ must be divisible by $6$ as $F_6 = 8$ and this means that the index of power of $2$ Fibonacci number greater that $8$ must be a power of 6 and therefore must be divisible by $F_{36}$.

However $F_{36}$ is also divisible by $F_{9}$ since $9$ divides $36$ and given that $F_9 = 34, F_{36}$ is therefore divisible by $34$ and cannot be a power of $2$.

Since any candidate powers of $2$ greater than $8$ must be divisible by $34$ there can be no Fibonacci numbers greater than $8$ which are powers of $2$.

As far as I can tell, as a corollary of Carmichael's theorem, it follows that no Fibonacci numbers other than $1$, $2$ and $8$ can be powers of $2$. Thus, $8$ is the largest.

There are three, and the biggest is 8.