Showing $\gcd(2^m-1,2^n+1)=1$

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 $(1)$.