I posted this conjecture on math.stackexchange, but I received no answer proving or disproving it: if $ m > 4 $ is a positive integer not divisible by $ 2 $ or $ 3 $, it's ever possible to find a positive integer $ n $ such that the difference of the $ 2 $ Fibonacci numbers $ F_{m+n}-F_n $ is a prime number.

Various example of $ [m, n] $, with the smallest n, are $[5,1],[7,3], [11,4], [13,5], [17,3],..., [619,1353],...,[2011,6629]...$. The numerical evidence supporting that is quite good; I need to thank @Peter for his great findings searching for counterexamples. Here you'll find all the values of $n$ for $m < 3000$ supporting the conjecture (the hardcases are at the bottom). Here the Oeis sequence about the difference of Fibonacci numbers being a prime

I converted the primes found that way in the Zeckendorf representation and they are an alternating sequence of $ 1 $ and $ 0 $ ending with two adjacent $ 0 $ and $ 1 $ (and eventually just various $ 0 $). Looking for some known result, I found this article: "Primes as Sums of Fibonacci Numbers", I asked about this conjecture to one of the authors and this was the answer:

"The conjecture can be stated as follows. First, for $ k≥1 $, we define $K$ to be the string of length $2k-1$, starting with $1$ and containing $1s$ and $0s$ alternatingly.Is there always a prime number $p$ having the Zeckendorf expansion $K0010..0$? This would imply the "showcase result" of our paper, "for each large enough $k$, there is a prime number $p$ that is the sum of exactly $k$ different, non-adjacent Fibonacci numbers". I don't see a way how to prove this, as it amounts to finding a prime number in a very sparse subsequence (growing exponentially). In our paper the problem was easier, since there are "many" integers whose Zeckendorf expansion has length $Ck$ ($C$ some constant involving the golden ratio) and exactly $k$ digits equal to $1$.

Do you have any idea?