Coordinates never increase, so the first axis point reached is $(0,0)$ exactly when the particle reaches $(1, 1)$ and then takes the diagonal step. Every path from $(4,4)$ to $(1,1)$ automatically stays off the axes, since its coordinates remain at least $1.$

A path from $(4,4)$ to $(1,1)$ with $d$ diagonal steps also has $3 - d$ left steps and $3 - d$ down steps, for $6 - d$ steps in all, and there are $\frac{(6-d)!}{d!\,(3-d)!\,(3-d)!}$ orderings: $20, 30, 12, 1$ for $d = 0, 1, 2, 3.$ Since a path with $6 - d$ steps has probability $\left(\frac{1}{3}\right)^{6-d},$ the probability of reaching $(1,1)$ and then stepping to $(0,0)$ is $$\begin{aligned} &\frac{1}{3}\left(\frac{20}{3^6} + \frac{30}{3^5} + \frac{12}{3^4} + \frac{1}{3^3}\right) \\ &= \frac{1}{3} \cdot \frac{20 + 90 + 108 + 27}{3^6} \\ &= \frac{245}{3^7}. \end{aligned}$$

Since $245 = 5 \cdot 7^2$ is not divisible by $3,$ we get $m + n = 245 + 7 = 252.$