It is clear that if $m=n$, then we must have $m=n=1$. If $m\ne n$, without loss of generality, let's assume $1 \le m < n \le 1981$. Now, because $$ \begin{align} (n^2 - mn-m^2)^2 &= ((n-m)^2 +mn-2m^2)^2\\ &= ((n-m)^2 +m(n-m)-m^2)^2\\ &=(m^2 - m(n-m) -(n-m)^2)^2 \end{align} $$ This means that if $(m, n)$ is a solution, $(n-m, m)$ must be a solution too. This descent process cannot go on forever. It terminates only when $m=n=1$. Hence all the pairs that satisfy the relationship can be obtained by the inverse transformation $(m, n)\implies (n, m+n)$: $$(1, 1)\implies (2, 1) \implies (3, 2) \implies (5, 3)\implies \cdots$$ So the components of all such pairs are Fibonacci numbers. The largest Fibonacci number less than 198 is $F_{16}=1597$. So the answer is $F_{15}^2 + F_{16}^2=\boxed{3524578}$