Let me share an argument with a gap.

Assume contrary. By rescaling the second polyhedron, say $P$, we can assume that the radius of the sphere stays the same as for the original polyhedron, say $Q$, but all edges get shorter.

Consider the central projection of the edges of $Q$ to the sphere. Note that they cut the sphere into cyclic polygons. Cyclic polygon maximize the area among all polygons with the same sides.

In general, the area is not maximal for polygons with at most the same sides --- this is the problem with my argument. For many cyclic polygons it is true --- further assume it is true for every facet of $Q$.

Now project centrally, old edges of $P$ to the sphere. We arrive at a contradiction, since the area of the sphere stays the same and all polygons are getting smaller.

"Yes" for the case $m=n=3$. I will assume that center of sphere lies in the first polyhedron, say $Q$, but it is straigthforward to remove this assumption.

Assume contrary. By rescaling the second polyhedron, say $P$, we can assume that the radius of the sphere stays the same as for the original polyhedron $Q$, but all edges get shorter.

Consider the central projection of the edges of $Q$ to the sphere. Note that they cut the sphere into cyclic polygons. Choose an edge $e$ of $Q$, consider two of its adjacent spherical cyclic polygons $F$ and $G$. The vertices of $F$ and $G$ lie on two arcs with endpoints of $e$. Since $Q$ is convex the arcs bound a nonconvex region from the sphere with $F$ and $G$ inside. The last statement (with some work) implies the following: if one decreases the edge $e$ while keeping $F$ and $G$ cyclic, then the total area of $F$ and $G$ decreases.

Now, for each edge of $Q$ draw a line segment between the corresponding vertices of $P$. Project the obtained line segments to the sphere. Since the center of sphere is inside of $P$ the spherical polygons that correspond to facets of $Q$ will cover the whole sphere. Note that cyclic polygon maximize the area among all polygons with the same sides. From above it follows that the total sum of their areas get smaller, but they cover the same sphere --- a contradiction.

Let $q_1,\dots,q_s$ be the vertices of $P$; denote by $o$ the center of $S$.

Note that $q_i\mapsto p_i$ is a distance-nonexpanding map. By Kirszbraun theorem it can be extended to a distance-nonexpanding map $f$ defined on the whole space. It follows that $p_1,\dots,p_s$ lie on the distance $\le r$ from $f(o)$. So the radius of $\{p_1,\dotsc,p_s\}$ is at most $r$.

Since the circumcenter of $\{p_1,\dotsc,p_s\}$ lies in $\operatorname{conv}\{p_1,\dotsc,p_s\}$, the circumradius $=$ radius of $\{p_1,\dotsc,p_s\}$. Whence the statement follows.

Let me share an argument with a gap.

Assume contrary. By rescaling the second polyhedron, say $P$, we can assume that the radius of the sphere stays the same as for the original polyhedron, say $Q$, but all edges get shorter.

Consider the central projection of the edges of $Q$ to the sphere. Note that they cut the sphere into cyclic polygons. Cyclic polygon maximize the area among all polygons with the same sides.

In general, the area is not maximal for polygons with at most the same sides --- this is the problem with my argument. For many cyclic polygons it is true --- further assume it is true for every facet of $Q$.

Now project centrally, old edges of $P$ to the sphere. We arrive at a contradiction, since the area of the sphere stays the same and all polygons are getting smaller.