TITLE: Can the difference of two distinct Fibonacci numbers be a square infinitely often?
QUESTION [32 upvotes]: Can the difference of two distinct Fibonacci numbers be a square infinitely often?

There are few solutions with indices $<10^{4}$ the largest two being $F_{14}-F_{13}=12^2$ and $F_{13}-F_{11}=12^2$
Probably this means there are no identities between near neighbours.
Since Fibonacci numbers are the only integral points on some genus 0 curves the problem is equivalent to finding integral points on one of few varieties. Fixing $F_j$ leads to finding integral points on a quartic model of an elliptic curve.

Are there other solutions besides the small ones?

[Added later] Here is a link to elliptic curves per  François Brunault's comment.
According to DIOPHANTINE EQUATIONS, FIBONACCI HYPERBOLAS,. AND QUADRATIC FORMS. Keith Brandt and John Koelzer.
Fibonacci numbers with consecutive odd indices are the only solutions to
$$ x^2-3xy+y^2 = -1 \qquad (1)$$
Fibonacci numbers with consecutive even indices are the only solutions to
$$ x^2-3xy+y^2 = 1 \qquad (2)$$
Given $F_n$ and $F_{n+2}$ one can compute $F_{n+k}$ using the linear Fibonacci recurrence and $F_{n+k}$ will be a linear combination $l(x,y)$ of $x,y$. Adding $l(x,y)-x=z^2$ to (1) or (2) gives a genus 1 curve. (Or just solve $l(x,y)-x=z^2$ and substitute in (1) or (2) to get a genus 1 quartic).
The closed form of $l(x,y)$ might be of interest, can't find the identity at the moment.
Probably a genus 0 curve with integral points $F_{2n},F_{2n+1}$ will be better.
Added much later

Does some generalization of the $abc$ conjecture predict something?

For 3 Fibonacci numbers identities are much easier:
$$  F_{4n+1}+F_{4n+3}-F_3 = L_{2n+1}^2$$

REPLY [7 votes]: This is a slight elaboration on joro's comment; I was hoping that someone else would write a better version of this.
The integer points on $x^2-xy-y^2 = 1$ are precisely the pairs $(F_{2n+1}, F_{2n})$. So looking for solutions of the form $F_{2n+1} - F_{2m+1} = z^2$ is looking for integer points on $$ x_1^2 - x_1 y_1 - y_1^2 = x_2^2 - x_2 y_2 - y_2^2 = 1,\ x_1 -x_2 = z^2.$$
The other three possibilities give similar equations. 
Each of these is a $K3$ surface. Here is where a better answer would review the major results on integer points on $K3$ surfaces. But I don't know them, so I'm going to stop here and hope someone else fills it in.