Given two integers $m$ and $n$ satisfying $4 < m < n$, let $A_1A_2\cdots A_{2n+1}$ be a regular $2n+1$ polygon. Denote by $P$ the set of its vertices. We aim to find the number of convex $m$-gons whose vertices belong to $P$ and have exactly two acute angles.

Notice that if a regular $m$-gon has exactly two acute angles, they must be at consecutive vertices. Otherwise, there would be two disjoint pairs of sides that take up more than half of the circle each. Assume that the last vertex, clockwise, of these four vertices that make up two acute angles is fixed; this reduces the total number of regular $m$-gons by a factor of $2n + 1$, and we will later multiply by this factor.

Suppose the larger arc that the first and the last of these four vertices make contains $k$ points, and the other arc contains $2n - 1 - k$ points. For each $k$, the vertices of the $m$-gon on the smaller arc may be arranged in $\binom{2n - 1 - k}{m - 4}$ ways, and the two vertices on the larger arc may be arranged in $(k - n)^2$ ways (so that the two angles cut off more than half of the circle).

The total number of polygons given by $k$ is thus $(k - n)^2 \times \binom{2n - 1 - k}{m - 4}$. Summation over all $k$ and change of variable gives that the total number of polygons (divided by a factor of $2n + 1$) is:
$$\sum_{k \geq 0} k^2 \binom{n - k - 1}{m - 4}.$$

This can be proven to be exactly $\binom{n}{m - 1} + \binom{n + 1}{m - 1}$ by double induction on $n > m$ and $m > 4$. The base cases $n = m + 1$ and $m = 5$ are readily calculated. The induction step is:
$$\sum_{k \geq 0} k^2 \binom{n - k - 1}{m - 4} = \sum_{k \geq 0} k^2 \binom{(n - 1) - k - 1}{m - 4} + \sum_{k \geq 0} k^2 \binom{(n - 1) - k - 1}{(m - 1) - 4}.$$
$$= \binom{n - 1}{m - 1} + \binom{n}{m - 1} + \binom{n - 1}{m - 2} + \binom{n}{m - 2} = \binom{n}{m - 1} + \binom{n + 1}{m - 1}.$$

So the total number of $m$-gons is:
$$(2n + 1) \times \left[ \binom{n}{m - 1} + \binom{n + 1}{m - 1} \right].$$

The answer is: \boxed{(2n + 1) \left[ \binom{n}{m - 1} + \binom{n + 1}{m - 1} \right]}.