58th Ukrainian National Mathematical Olympiad

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

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