Put the origin at the point where the axes cross, and measure signed coordinates $u_1, u_2$ along the two axes. Along its axis, each cone has its apex at distance $8 - 3 = 5$ from the origin and its base plane at distance $3$ on the other side. Slicing a cone by any plane through its axis gives a triangle whose slant side, in coordinates $(u, w)$ with $w \ge 0$ the distance from the axis, is the line through $(5, 0)$ and $(-3, 3),$ namely $3u + 8w = 15.$

A sphere of radius $r$ centered at a point with axial coordinate $u$ and distance $\rho$ from the axis fits inside that cone only if its cross-section fits inside the triangle, so $r \le \frac{15 - 3u - 8\rho}{\sqrt{73}}.$ Since the two axes are perpendicular, the distance from the center to axis $1$ is at least $|u_2|,$ and vice versa. Adding the two constraints, $$\begin{aligned} 2\sqrt{73}\,r &\le 30 - 3(u_1 + u_2) \\ &\quad {}- 8(|u_1| + |u_2|) \\ &\le 30, \end{aligned}$$ because $3(u_1 + u_2) \ge -3(|u_1| + |u_2|)$ $\ge -8(|u_1| + |u_2|).$ Hence $r \le \frac{15}{\sqrt{73}}.$

The sphere of radius $\frac{15}{\sqrt{73}}$ centered at the origin achieves this: its distance to each slant surface is $\frac{|3 \cdot 0 + 8 \cdot 0 - 15|}{\sqrt{3^2 + 8^2}} = \frac{15}{\sqrt{73}},$ and its distance $3$ to each base plane is larger. So the maximum of $r^2$ is $\frac{225}{73},$ and $m + n = 225 + 73 = 298.$