Ukrainian National Mathematical Olympiad

If $m = n$, then $A = 2m^{525}$, $B = 2m \cdot 2m^2 \cdot 2m^4 \dots \cdot 2m^{128} = 2^8 m^{255} > A = 2m^{525}$
$\Leftrightarrow m = n = 1$, while for $m = n > 1$ we get $A > B$.

Now, without the loss of generality, $m > n$. Then we can do the transformations:
$$\begin{aligned} B \le (m-n)B &= (m-n)(m+n)(m^2+n^2)(m^4+n^4)\dots(m^{128}+n^{128}) = \\ &= (m^2-n^2)(m^2+n^2)(m^4+n^4)\dots(m^{128}+n^{128}) = (m^4-n^4)(m^4+n^4)\dots(m^{128}+n^{128}) = \\ &= m^{256}-n^{256} < m^{525} + n^{525} = A \Rightarrow B < A. \end{aligned}$$