Determine the maximum value of $m^{2}+n^{2}$, where $m$ and $n$ are integers satisfying $m,n\in \{1,2,...,1981\}$ and $(n^{2}-mn-m^{2})^{2}=1.$
To determine the maximum value of \( m^2 + n^2 \), where \( m \) and \( n \) are integers satisfying \( m, n \in \{1, 2, \ldots, 1981\} \) and \((n^2 - mn - m^2)^2 = 1\), we need to solve the equation \((n^2 - mn - m^2)^2 = 1\).

This equation can be rewritten as:
\[ n^2 - mn - m^2 = \pm 1 \]

We will consider both cases separately.

### Case 1: \( n^2 - mn - m^2 = 1 \)

1. Rewrite the equation:
   \[ n^2 - mn - m^2 + 1 = 0 \]

2. This is a quadratic equation in \( m \):
   \[ m^2 + mn - n^2 + 1 = 0 \]

3. The discriminant of this quadratic equation must be a perfect square:
   \[ \Delta = n^2 + 4(n^2 - 1) = 5n^2 - 4 \]

4. Set \( 5n^2 - 4 = k^2 \) for some integer \( k \):
   \[ 5n^2 - k^2 = 4 \]

5. This is a Pell's equation of the form \( x^2 - 5y^2 = 4 \). The fundamental solution is \( (x_0, y_0) = (3, 1) \).

6. The general solution to this Pell's equation is given by:
   \[ n_r = \frac{(3 + \sqrt{5})^r + (3 - \sqrt{5})^r}{2} \]
   \[ k_r = \frac{(3 + \sqrt{5})^r - (3 - \sqrt{5})^r}{2\sqrt{5}} \]

7. We need \( n \leq 1981 \). The sequence of \( n \) values is:
   \[ n_0 = 1, n_1 = 2, n_{r+2} = 3n_{r+1} - n_r \]

8. The corresponding \( k \) values are:
   \[ k_0 = 1, k_1 = 4, k_{r+2} = 3k_{r+1} - k_r \]

9. For \( n = 610 \), we get \( k = 1364 \) and \( m = \frac{n + k}{2} = \frac{610 + 1364}{2} = 987 \).

### Case 2: \( n^2 - mn - m^2 = -1 \)

1. Rewrite the equation:
   \[ n^2 - mn - m^2 - 1 = 0 \]

2. This is a quadratic equation in \( m \):
   \[ m^2 + mn - n^2 - 1 = 0 \]

3. The discriminant of this quadratic equation must be a perfect square:
   \[ \Delta = n^2 + 4(n^2 + 1) = 5n^2 + 4 \]

4. Set \( 5n^2 + 4 = k^2 \) for some integer \( k \):
   \[ k^2 - 5n^2 = 4 \]

5. This is another Pell's equation of the form \( x^2 - 5y^2 = 4 \). The fundamental solution is \( (x_0, y_0) = (3, 1) \).

6. The general solution to this Pell's equation is given by:
   \[ n_r = \frac{(3 + \sqrt{5})^r + (3 - \sqrt{5})^r}{2} \]
   \[ k_r = \frac{(3 + \sqrt{5})^r - (3 - \sqrt{5})^r}{2\sqrt{5}} \]

7. We need \( n \leq 1981 \). The sequence of \( n \) values is:
   \[ n_0 = 1, n_1 = 2, n_{r+2} = 3n_{r+1} - n_r \]

8. The corresponding \( k \) values are:
   \[ k_0 = 3, k_1 = 7, k_{r+2} = 3k_{r+1} - k_r \]

9. For \( n = 987 \), we get \( k = 2207 \) and \( m = \frac{n + k}{2} = \frac{987 + 2207}{2} = 1597 \).

### Conclusion

The maximum value of \( m^2 + n^2 \) is obtained by comparing the results from both cases:
- Case 1: \( m = 987, n = 610 \)
- Case 2: \( m = 1597, n = 987 \)

Calculate \( m^2 + n^2 \) for both cases:
- Case 1: \( 987^2 + 610^2 = 974169 + 372100 = 1346269 \)
- Case 2: \( 1597^2 + 987^2 = 2550409 + 974169 = 3524578 \)

Thus, the maximum value of \( m^2 + n^2 \) is \( 3524578 \).

The final answer is \( \boxed{3524578} \)