Problem 768

[Solution] Consider the distance between any two points in $M$, and let $A B$ be the line segment connecting the two points with the greatest distance.
By the problem's condition, there exists a point $C$ such that $\triangle A B C$ is an equilateral triangle.
Construct arcs with $A, B, C$ as centers and $A B$ as the radius, then we get the curved triangle as shown in the figure, and all points in $M$ are within this curved triangle.
If $M$ has another point $P$ different from $A, B, C$.
Consider the position of $P$.
(1) If $P$ is inside or on the boundary of $\triangle A B C$, then for $A, P$ there is a third point $P^{\prime}$ such that $\triangle A P P^{\prime}$ is an equilateral triangle. At this time, $P^{\prime}$ cannot be inside $\triangle A B C$, $P^{\prime}$ can only be inside or on the boundary of one of the segments of the arc.
Connect $C P^{\prime}$, then there is a third point $P^{\prime \prime}$ such that $\triangle C P^{\prime} P^{\prime \prime}$ is an equilateral triangle.
If $P^{\prime \prime}$ and $A$ are on the same side of the line $C P^{\prime}$, then by
$$\angle P^{\prime \prime} A C=\angle P^{\prime} B C>60^{\circ},$$

at this time $P^{\prime \prime}$ is outside the curved triangle.
If $P^{\prime \prime}$ and $B$ are on the same side of the line $C P^{\prime}$, then similarly, $P^{\prime \prime}$ is outside the curved triangle.
Thus, a contradiction arises.
(2) If $P$ is inside or on the boundary of the segment of the arc, by (1), $P^{\prime}$ is also outside the curved triangle, leading to a contradiction.
By (1) and (2), the maximum number of elements in $M$ is 3.