Are $1$ and $3$ the only numbers of the form $2^n-1$ that exist in the Fibonacci sequence?

Question

Are $1$ and $3$ the only numbers of the form $2^n-1$ that exist in the Fibonacci sequence?

Then, if they are not the only ones, are there infinite cases?

I have tried finding another example using a program, but could not find any, thus my conclusion.

I think there might be a good proof in the fact that all Fibonacci numbers $Fn$ create perfect squares when $5{Fn}^2+4$ or $5(Fn)^2-4$.