Solution:
The number $m - n$ has at most three different positive divisors if and only if $m - n = p^k$, where $p$ is a prime and $k \in \{0, 1, 2\}$. If $k = 0$ then $m = n + 1$ and $n(n + 1)$ is a perfect square, which is impossible.

Let $m - n = p^k$, $mn = t^2$, where $k \in \{1, 2\}$ and $t$ is a positive integer. Then
$$n\left(n + p^k\right) = t^2 \Longleftrightarrow \left(2n + p^k - 2t\right)\left(2n + p^k + 2t\right) = p^{2k}$$
Therefore $2n + p^k - 2t = p^s$ and $2n + p^k + 2t = p^r$, where $r$ and $s$ are integers, such that $0 \leq s < r \leq 2k$ and $r + s = 2k$.

For $k = 1$ we have the unique possibility $2n + p - 2t = 1$, $2n + p + 2t = p^2$. Hence
$$n = \frac{(p - 1)^2}{4}, \quad m = n + p = \frac{(p + 1)^2}{4}$$
Taking into account that $1000 \leq m < 2005$ we obtain the solutions $m = 1764$, $1600$, $1369$, $1296$ and $1156$ (for $p = 83, 79, 73, 71$ and $67$, respectively).

For $k = 2$ we have $(r, s) = (4, 0)$ or $(r, s) = (3, 1)$. In the first case we get
$$n = \frac{\left(p^2 - 1\right)^2}{4}, \quad m = n + p^2 = \frac{\left(p^2 + 1\right)^2}{4}$$
Now the inequalities $1000 \leq m < 2005$ imply that $p = 8$ which is not a prime. In the second case we have $m = p(p + 1)^2 / 4$, which gives the solutions $m = 1900$ and $1377$ (for $p = 19$ and $17$, respectively).