We can analyze that we can eliminate the product term, thereby observing a connection to Pell's equation.
Solve $\left(n^{2}+2 m n-2 m^{2}\right)^{2}=\left[(n+m)^{2}-3 m^{2}\right]^{2}=1$, so $(n+m)^{2}-3 m^{2}= \pm 1$. Let $x=m+n$, $y=n$, then $x^{2}-3 y^{2}= \pm 1$. For $x^{2}-3 y^{2}=-1$, taking both sides modulo 3, we find that the equation has no integer solutions.

Therefore, $x^{2}-3 y^{2}=1$. Its smallest positive integer solution is $x_{1}=2, y_{1}=1$, and by the theorem of Pell's equation, we get $x_{2}=7$, $y_{2}=4$. If all solutions are $\left\{x_{n}, y_{n}\right\}\left(n \in \mathbf{N}^{*}\right)$, then the second-order linear recurrence relation for $\left\{x_{n}\right\}$ is $x_{n+1}=4 x_{n}-x_{n-1}$; the second-order linear recurrence relation for $\left\{y_{n}\right\}$ is $y_{n+1}=4 y_{n}-y_{n-1}$. To maximize $m^{2}+n^{2}$, we need $m, n$ to take their maximum values. From $x=m+n, y=n$, we get $m=x-y, n=y$. It is clear that when $x, y$ take their maximum values, $m, n$ also have their maximum values. By the recurrence relation, we calculate:
$$
y_{6}=780, y_{7}=2911>2005,
$$

Thus, when $y_{6}=780, x_{6}=1086$, $m, n$ have their maximum values, $\left\{\begin{array}{l}m=306 \\ n=780\end{array}\right.$.

In summary, when $m=306, n=780$, $m^{2}+n^{2}=702036$ is the maximum value that satisfies the given conditions.
For example, $\mathbb{S}$ determine the maximum value of $m^{2}+n^{2}$, where $m, n \in\{1,2, \cdots, 2005\}$ satisfy $\left(n^{2}+2 m n-2 m^{2}\right)^{2}=1$.

