Solution:

In an equiangular pentagon each side direction turns by the exterior angle $72^\circ,$ so the sides are the vectors $s_k u_k$ for $k = 1, \ldots, 5,$ where $u_k = (\cos 72k^\circ, \sin 72k^\circ)$ and $\sum_k s_k u_k = 0.$ Write $Q = \sum s_k^2 = 308,$ $P_1 = \sum_{k} s_k s_{k+1},$ and $P_2 = \sum_k s_k s_{k+2}$ (indices cyclic). Each diagonal is a sum of two consecutive side vectors, so its square is $s_{k+1}^2 + s_{k+2}^2 + 2 s_{k+1} s_{k+2} \cos 72^\circ,$ and summing all five gives $$800 = 2Q + 2\cos 72^\circ \, P_1, \qquad \text{so} \qquad 2\cos 72^\circ \, P_1 = 800 - 616 = 184.$$

Expanding $\left|\sum_k s_k u_k\right|^2 = 0,$ the angle between $u_k$ and $u_{k+1}$ is $72^\circ$ and between $u_k$ and $u_{k+2}$ is $144^\circ:$ $$0 = Q + 2\cos 72^\circ \, P_1 + 2\cos 144^\circ \, P_2 = 308 + 184 - 2\cos 36^\circ \, P_2,$$ so $2\cos 36^\circ \, P_2 = 492.$ Using $\cos 72^\circ = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4},$ we get $2P_1 = \frac{184}{\cos 72^\circ} = 184\left(\sqrt{5} + 1\right)$ and $2P_2 = \frac{492}{\cos 36^\circ} = 492\left(\sqrt{5} - 1\right).$

The square of the perimeter is $$\left(\sum s_k\right)^2 = Q + 2P_1 + 2P_2 = 308 + 184\sqrt{5} + 184 + 492\sqrt{5} - 492 = 676\sqrt{5}.$$ Therefore $m + n = 676 + 5 = 681.$