If $n > m+1$, then by Pigeonhole there exists at least two sides of the $m$-gon that contains two points of the $n$-gon, as having a side of the $m$-gon contain three or more points of the $n$-gon is clearly preposterous. By symmetry, the two points on each side of the $m$-gon have equal distance from the midpoint of that side. If the two sides of the $m$-gon that each have two vertices of the $n$-gon are not diametrically opposite, then the center of the $n$-gon is the intersection of the perpendicular bisectors of its two sides lying on the two sides of the $m$-gon, which is precisely the center of the $m$-gon itself. Thus, drawing the circumcircle of the $n$-gon, it intersects each of the $m$-gon's sides at two points, so $n = 2m$. If the two sides of the $m$-gon that each have two points of the $n$-gon are diametrically opposite and all other sides of the $m$-gon each have one side of the $n$-gon (it is clear that there cannot be a side of the $m$-gon with zero points of the $n$-gon, else the $n$-gon's side length is larger than the $m$-gon, making it impossible to inscribe), then it follows that the circumcircle of the $n$-gon is tangent to all sides of the $m$-gon except the two diametrically opposite sides with two points each. But for $m \ge 6$, this circle is the incircle of at least three sides and is thus uniquely determined, and as the incircle of the $m$-gon satisfies this, it must be the only possible circumcircle of the $n$-gon, contradiction. For $m=4$, it is easy to see that we cannot inscribe a hexagon in a square. It remains to solve the case $n=m+1$. In this case, we must have exactly one side of the $m$-gon with two vertices of the $n$-gon, with all other sides having one vertex. Clearly $n=4, m=3$ is possible. Otherwise, $m\ge 4$. Label $V_1, V_2, \ldots , V_m$ the vertices of the $m$-gon, and $X_1, X_2, \ldots , X_{m+1}$ the vertices of the $n$-gon, with $X_1, X_2 \in V_1V_2$, and $X_i \in V_{i-1}V_i$ for $3\le i\le m+1$, where $V_{m+1}=V_1$. Since $\angle X_2V_2X_3 = \angle X_3V_3X_4 = \angle X_4V_4X_5$ and $X_2X_3=X_3X_4=X_4X_5$, then the circumcircles $(X_2V_2X_3), (X_3V_3X_4) ,(X_4V_4X_5)$ are congruent. Let $\{P_1, X_3\} = (X_2V_2X_3)\cap (X_3V_3X_4)$ and $\{P_2, X_4\} = (X_3V_3X_4)\cap (X_4V_4X_5)$. Then by spiral similarity along with the circle congruency and $V_2V_3=V_3V_4$, $\triangle P_1V_2V_3\cong \triangle P_2V_3V_4$. Thus, either $\angle X_3X_2V_2=\angle X_4X_3V_3 \implies \angle X_2X_3V_2=0$ clearly preposterous, or $\angle X_3X_2V_2 = \angle X_3X_4V_3\implies \angle X_5X_4V_4 = 0$, again preposterous. (Note that this argument fails for $n=4, m=3$ since that configuration does not contain enough vertices.) Thus, our only answers are $\boxed{(m, n) = (m, 2m), (3, 4)}$.