Showing $\gcd(2^m-1,2^n+1)=1$ A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement 

If $m$ is odd then $\gcd(2^m-1,2^n+1)=1$. 

It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues.  She has not taken any group theory.
 A: If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p \neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$\gcd (2^t-1, 2^u-1) = 2^{\gcd (t,u)}-1\tag{1}$$
to conclude
$$\gcd (2^m-1, 2^{2n}-1) = 2^{\gcd(m,2n)}-1.$$
But since $m$ is odd, we have $\gcd (m,2n) = \gcd(m,n)$, and hence 
$$2^{\gcd(m,2n)}-1 \mid 2^n-1,$$
which, since
$$\gcd(2^n-1,2^n+1) = \gcd(2^n-1,2) \mid 2$$
and $2^{\gcd(m,2n)}-1$ is odd, implies $\gcd (2^{\gcd(m,2n)}-1,2^n+1) = 1$ and hence $\gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = q\cdot t + r$ with $0 \leqslant r < t$, and
$$2^u-1 = 2^r\left(2^{q\cdot t}-1\right) + \left(2^r-1\right),$$
which, since $2^t-1 \mid (2^t)^q-1$, yields
$$\gcd(2^t-1,2^u-1) = \gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
A: First thing i did was $2^m+1=2^{m-n}\cdot 2^{n}+2^{m-n}-2^{m-n}+1=2^{m-n}(2^n-1)+(2^{m-n}+1)$
which lead me to $\gcd(2^{m}+1,2^{n}-1)=\gcd(2^{m-n}+1,2^{n}-1)$
continuing till difference between $m$ and $n$ become $1$
finaly I obtained is $\gcd(2^1+1,2^n-1)$
which is $\gcd(3,2^{n}-1)=1,\quad$because n is odd.
