Example 1 The figure $F$ is enclosed by the line $y=x$ and the parabola $y=2^{-100} x^{2}$. Then, how many points of the form $\left(2^{m}, 2^{n}\right)$ are there inside the figure $F$ (excluding the boundary) $\left(m, n \in \mathbf{N}_{+}\right)$.
Solve $y=x$ and $y=2^{-100} x^{2}$ intersect at $(0,0)$ and $\left(2^{100}, 2^{100}\right)$, the point $\left(2^{m}, 2^{n}\right)$ should satisfy $0$ $n>2 m-100$, when $m=k$ ( $2 \leqslant k \leqslant 50)$, $n$ can take $1,2, \cdots, k-1$, when $m=k(51 \leqslant k \leqslant 99)$, $n$ can take $2 k-99,2 k-98, \cdots, k-1$, so there are
$$
\begin{aligned}
\sum_{k=2}^{50}(k-1)+\sum_{k=51}^{99}(99-k) & =\frac{50 \times 49}{2}+\frac{49 \times 48}{2} \\
& =2401 \text { points of the form }\left(2^{m}, 2^{n}\right) \text { (where }m, n \in \mathbf{N}_{+}\text{) }.
\end{aligned}
$$