We want $2^x + 2^{2y} + 2^{3z} + 2{8} $to be a power of$ 2$. We must have two of the powers equal to each other otherwise we could bring out the lowest factor of two and the remaining part would be odd. Applying the same logic repeatedly and caseworking we get the only solution as $(x, y, z) = (10 , 4 , 3)$

(x,y,z) = (8,5,3) and (7,3,2) works as well