One solution may lead to another one.

Experimenting with small values suggests that the solutions of $n^{2}$ - $mn$ - $m^{2}$ = 1 or -1 are successive Fibonacci numbers.

So suppose $n > m$ is a solution.

This suggests trying m+n, n:

$(m+n)^{2}$ - $(m+n)n$ - $n^{2}$ = $m^{2}$ + $mn$ - $n^{2}$ = -( $n^{2}$ - $mn$ - $m^{2}$ ) = 1 or -1.

So if n > m is a solution, then m+n, n is another solution. Running this forward from 2,1 gives 3,2; 5,3; 8,5; 13,8; 21,13; 34,21; 55,34; 89,55; 144,89; 233,144; 377,233; 610,377; 987,610; 1597,987; 2584,1597.

But how do we know that there are no other solutions????????????????

The trick is to run the recurrence the other way. For suppose n > m is a solution, then try m, n-m:

$m^{2}$ - $m(n-m)$ - $(n-m)^{2}$ = $m^{2}$ + $mn$ - $n^{2}$ = -( $n^{2}$ - $mn$ - $m^{2}$ ) = 1 or -1,

so that also satisfies the equation. Also if m > 1, then m > n-m (for if not, then n >= 2m, so n(n - m) >= $2m^{2}$ , so $n^{2}$ - $nm$ - $m^{2}$ >= $m^{2}$ > 1). So given a solution n > m with m > 1, we have a smaller solution m > n-m. This process must eventually terminate, so it must finish at a solution n, 1 with n > 1. But the only such solution is 2, 1. Hence the starting solution must have been in the forward sequence from 2, 1.

Hence the solution to the problem stated is $1597^{2}$ + $987^{2}$ = $\boxed{3524578}$

Heres a general pattern in n and m values i.e (1,1),(2,1),(3.2),.......(1597,987) .Its a sequence with(f(n+1),f(n)) where f(n) is a fibonacci sequence