If $m,n\in\mathbb{N}$ and $f(m) \mid f(n)$ for the $f$ given below, must $m=n$? If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?

False as stated. The divisibility also holds for example, when $n=1$ and $m=13$. 
When $m=1$, the left hand side is
$$
(2^3-1)2^2(2^3+2^2+1)=7\cdot4\cdot13.
$$
When $n=13$ we obviously have that
$$
(2^3-1)\mid(2^{27}-1)\qquad\text{and}\qquad2^2\mid2^{4\cdot13-2}.
$$
The last factor is a bit trickier, but from Little Fermat we see that modulo $13$ we have the congruences $2^{13+1}\equiv2^{1+1}$ and $2^{2\cdot13+1}\equiv2^{2\cdot1+1}$, so also
$$
2^{27}+2^{14}+1\equiv2^3+2^2+1\equiv0\pmod{13}.
$$
It seems to me that is should be easy to find several other examples of `factor wise' divisibility.
