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If you're allowing negative numbers, then $F_{36} - F_{-12} = 3864^2$. This comes from writing $F_{n + 4m} - F_n = F_{2m}L_{n+2m}$ and splitting $F_{2m} = F_mL_m$, then taking $F_m$ to be a square and setting $m= n+2m$. As before, this only works because $F_m$ is a square, which only happens a few times. Also note, allowing negatives is the same as asking "When is $F_{2n + m} + F_{2n}$ a square?", since we only duplicate our positive solutions if $F_{-n} = F_n$.
In a similar fashion, you can show that there is a solution (and exactly one) separated by 6 terms: $F_{15} - F_9 = 576 = 24^2$. But this example relies on the Lucas number $L_3$ being a square, so there's no hope that it produces an infinite family.
Using Cohn's characterization of when $F_n = 2x^2$ (it's true only when n = 0, 3, or 6), and simplifying $F_{n+3} - F_n = 2F_{n+1}$, the only examples separated by 3 are $F_5 - F_2 = 4$, and $F_8 - F_5 = 16$. Similarly, to find all pairs separated by 4, simplify $F_{n+2} - F_{n-2} = F_{n+1} + F_{n-1} = L_n$ to get that the Lucas number $L_n$ must be a square, so n = 1 or 3, giving $F_5 - F_1 = 4$ as the only solution. Both characterizations come from J. Cohn's paper "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 1964, pp. 109-113.
Winning Ways is fantastic for combinatorial number theory, because everything is built up through games which anyone can play, and numbers only come in as a notation to keep track of who is ahead in the game. Plus the "numbers" include surreal numbers like stars and arrows. I remember seeing this during a math summer camp when I was 15 and being shocked that something that wasn't a number (like "double up star") could be used to count things, and that those numbers could be combined (added) in a meaningful way.
Harold Edwards, in "Fermat's Last Theorem" (Spring-Verlag, 1977) also notes that Euler's original proof via descent fails for the above reasons, and gives a proof that cubes must factor into cubes by using methods known to Euler (arguments Euler used to show that primes congruent to 1 mod 3 can be written as $a^2 + 3b^2$), but also explicitly states that though Euler could have made this argument, he did not.