Answer: 113

Solution:

Let $y = 2^{111x}.$ Then $2^{333x-2} = \frac{y^3}{4},$ $2^{111x+2} = 4y,$ and $2^{222x+1} = 2y^2,$ so the equation becomes $\frac{y^3}{4} + 4y = 2y^2 + 1,$ that is, $$y^3 - 8y^2 + 16y - 4 = 0.$$ Since the three roots $x_1, x_2, x_3$ are real and $y = 2^{111x}$ is strictly increasing, they correspond to three positive real roots $y_1, y_2, y_3$ of the cubic.

Each $x_i = \frac{1}{111}\log_2 y_i,$ so $$\begin{aligned} x_1 + x_2 + x_3 &= \frac{1}{111}\log_2(y_1 y_2 y_3) \\ &= \frac{1}{111}\log_2 4 \\ &= \frac{2}{111}, \end{aligned}$$ using Vieta's formulas for the product of the roots. Then $m + n = 2 + 111 = 113.$