Answer: 77

Solution:

The first equation says $z$ lies on the circle of radius $5$ centered at $(25, 20).$ The second says $z$ is equidistant from $P_1 = (k + 4, 0)$ and $P_2 = (k, 3),$ i.e. it lies on the perpendicular bisector of $\overline{P_1 P_2}.$ The system has exactly one solution precisely when this line is tangent to the circle.

The midpoint is $\left(k + 2, \frac{3}{2}\right)$ and $\overline{P_1 P_2}$ has slope $-\frac{3}{4},$ so the bisector has slope $\frac{4}{3}:$ in standard form $8x - 6y - (8k + 7) = 0.$ Tangency requires $$\begin{gathered} \frac{|8 \cdot 25 - 6 \cdot 20 - 8k - 7|}{\sqrt{8^2 + 6^2}} \\ = \frac{|73 - 8k|}{10} \\ = 5, \end{gathered}$$ so $8k = 73 \pm 50,$ giving $k = \frac{123}{8}$ or $k = \frac{23}{8}.$

The sum is $\frac{146}{8} = \frac{73}{4},$ so $m + n = 73 + 4 = 77.$