Solution:

Adding $xyz$ to each equation gives $x^3 = 2 + xyz,$ $y^3 = 6 + xyz,$ and $z^3 = 20 + xyz.$ Let $P = xyz.$ Multiplying the three equations yields $$P^3 = (2 + P)(6 + P)(20 + P) = P^3 + 28P^2 + 172P + 240,$$ so $28P^2 + 172P + 240 = 0,$ i.e. $7P^2 + 43P + 60 = 0,$ whose roots are $P = -\frac{15}{7}$ and $P = -4.$ Each root is achievable: the cube roots of $2 + P,$ $6 + P,$ $20 + P$ then really do have product $P.$

Adding the original equations, $x^3 + y^3 + z^3 = 28 + 3P,$ which is maximized by the larger root $P = -\frac{15}{7}:$ $$a^3 + b^3 + c^3 = 28 - \frac{45}{7} = \frac{151}{7}.$$ Thus $m + n = 151 + 7 = 158.$