Show that for a given $s$ there are a finite number of Fibonacci number of form $n^2+s$

Pretty Hard Question: Partial Answer

$For\quad the\quad case\quad when\quad F_{ n }=m^{ 2 }+1,\quad F_{ n }-1=m^{ 2 }$

I. $n=4k\quad F_{4k}-1=F_{2k+1}L_{2k-1}$.

Note $L_{2k-1}=F_{2k+1}-F_{2k-3}$ where $gcd(F_{2k+1},F_{2k-3})=1$ Therefore, $L_{2k-1} and F_{2k+1}$ are relatively prime, and both must be squares. However, $L_{n}$ and $F_{n}$ both have only a finite number of solutions when they are square. Our proof is done.

II. $n=4k+1\quad F_{4k+1}-1=F_{2k}L_{2k+1}$.

Note $L_{2k+1}=F_{2k}+F_{2k+2}$ where $gcd(F_{2k+2},F_{2k})=1$. Therefore, $L_{2k+1} and F_{2k}$ are relatively prime, and both must be squares. However, $L_{n}$ and $F_{n}$ both have only a finite number of solutions when they are square. Our proof is done.

While I haven't been able to try the rest, I think they would follow a similar pattern.

Formula Used Presented in

Fibonacci[n]-1 is always composite for n>6. why?

https://en.wikipedia.org/wiki/Lucas_number