Show that for a given $s$ there are a finite number of Fibonacci number of form $n^2+s$ It is well known that the last Fibonacci number $F_k$ such that $\exists \ n \in \Bbb{N} : F_k = n^2$ is $144$.  Thus there are only $4$ perfect squares among the Fibonacci sequence (assuming  you count $F_0 = 0$ and count the duplicated $F_1 = F_2 = 1$).
The equation 
$$F_k = n^2 + 4
$$
also has a few solutions ($F_5 = 1^2 + 4, F_6 = 2^2 +4, f_7 = 3^2+4$) and appears to have no other solutions (although my proof of that might well have some holes).
Pretty Hard Question
Prove that 
$$F_k = n^2 + 1
$$
and 
$$F_k = n^2 -1
$$
each have only a finite number of solutions for $k,n \in \Bbb{N}$.
Challenging Question
Are there any values of $s$ such that
$$F_k = n^2 +s
$$
has no solutions for $k,n \in \Bbb{N}$?
Very Challenging Question:
Prove that for any $s \in \Bbb{Z}$ 
$$F_k = n^2 + s
$$
has only a finite number of solutions for $k,n \in \Bbb{N}$.
(Or find a counterexample $s$.)
 A: 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
A: The pretty hard question is completely resolved in this paper. The authors show that the only perfect powers (not only perfect squares!) which are at distance $1$ from a Fibonacci number are $0,1,2,3,5,8$.
