Proving the Fibonacci identity $(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$

I will give an outline of a proof, but leave the details to you. This is almost certainly not the method that you are expected to take, but it avoids the messy algebra of Binet's Formula.

One first proves (however they wish, perhaps by induction) that $$ F_{p+q} = F_{p-1}F_{q} + F_{p}F_{q+1} $$ for all integers $p,q$. Importantly, $p,q$ can be negative in the above formula. We also have that $$ F_{-n} = (-1)^{n+1}F_{n} $$ for all $n$. Then one gets $\textit{D'Ocagne's Identity}$: $$ F_{p-q} = (-1)^{q}\left(F_{p+1}F_{q-1} - F_{p-1}F_{q+1}\right) $$ as a simple corollary. Then from these one can prove that $$ F_{r}F_{a+b} = F_{a}F_{b+r} - (-1)^{r}F_{a-r}F_{b} $$ for all integers $a,b,r$. We can rewrite this as $$ F_{r}F_{a+b+r} = F_{a+r}F_{b+r} - (-1)^{r}F_{a}F_{b} $$ and hence if $a,b,c,d$ are integers such that $a+b=c+d$ then $$ F_{a+r}F_{b+r} - (-1)^{r}F_{a}F_{b} = F_{r}F_{a+b+r} = F_{r}F_{c+d+r} = F_{c+r}F_{d+r} - (-1)^{r}F_{c}F_{d} $$ and so for all such $a,b,c,d$ and any $r$ we have $$ F_{a}F_{b} - F_{c}F_{d} = (-1)^{r}\left(F_{a+r}F_{b+r} - F_{c+r}F_{d+r}\right) $$ From this we conclude that the LHS of your equation is simply $F_{2k+1}$, but then using the very first formula we wrote down we see that $$ F_{2k+1} = F_{(k+1) + k} = F_{k}^{2} + F_{k+1}^{2} $$

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Just to add a few comments. In your question you stated some restrictions on the values that $m,k$ can take. It is clear from my proof that in fact the formula remains true for any integers $m,k$. I suspect that the exercise is stated as it is to avoid considering Fibonacci numbers with negative coefficients, but really this is a natural consideration. Since most of these formulae that hold for the positive index Fibonacci numbers hold for the negative index Fibonacci numbers too.

Using Binet's formula, we have:

$F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$ where $\phi = \frac{1+\sqrt{5}}{2}, \psi = \phi-1 = \frac{-1}{\phi}$

Then subbing it into your equation:

$(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k})$

$ = (−1)^{m−k}(\frac{\phi^{m+k+1} - \psi^{m+k+1}}{\sqrt{5}} \frac{\phi^{m-k-1} - \psi^{m-k-1}}{\sqrt{5}} - \frac{\phi^{m+k} - \psi^{m+k}}{\sqrt{5}} \frac{\phi^{m-k} - \psi^{m-k}}{\sqrt{5}})$

$ = (−1)^{m−k}(\frac{\phi^{2m} + \psi^{2m} - \phi^{m+k+1}\psi^{m-k-1} - \psi^{m+k+1}\phi^{m-k-1}}{5} - \frac{\phi^{2m} + \psi^{2m} - \phi^{m+k}\psi^{m-k} - \psi^{m+k}\phi^{m-k}}{5})$

$ = (−1)^{m−k}(\frac{ - \phi^{m+k+1}\psi^{m-k-1} - \psi^{m+k+1}\phi^{m-k-1}}{5} - \frac{ - \phi^{m+k}\psi^{m-k} - \psi^{m+k}\phi^{m-k}}{5})$

$ = (−1)^{m−k}(\phi\psi)^{m+k}(\frac{ - \phi\psi^{-2k-1} - \psi\phi^{-2k-1}}{5} + \frac{ \psi^{-2k} + \phi^{-2k}}{5})$

$ = (−1)^{m−k}(\phi\psi)^{m+k}(\phi-\psi)(\frac{\psi^{-2k-1} - \phi^{-2k-1}}{5})$

$ = (−1)^{m−k}(-1)^{m+k}(\frac{\phi-\psi}{\sqrt{5}})(\frac{\psi^{-2k-1} - \phi^{-2k-1}}{\sqrt{5}})$

$ = (−1)^{2m}(F_1)(F_{-2k-1})$

Now using the fact that $F_{-n} = (-1)^{n+1}F_n$,

$(F_1)(F_{-2k-1}) = (-1)^{-2k}(F_{2k+1})$

Using some tedious algebra, one can also show that $F_{2k+1} = F_{k}^2 + F_{k+1}^2$ and so,

$(F_{2k+1}) = F_{k}^2 + F_{k+1}^2$