Fix $a,b>1$. Let $n$ be the smallest positive integer such that $a\mid 2^n-1$ and $b\mid 2^n+1$. Assume the contrary and take the smallest such $k$. Clearly $k\ne n$ as otherwise $a=b=1$.
Case 1: $k>n$. Let $k=qn+r$ for $0\le r\le n-1$. Note that $b\mid 2^k-1$ yields $2^n+1\equiv 2^r+1\pmod{b}$. Likewise, $2^n-1\equiv (-1)^q 2^r+1\pmod{a}$. If $q$ is odd, we get $a\mid 2^r-1$ and $b\mid 2^r+1$. Clearly $r\ne 0$, so $1\le r\le n-1$. This, however, contradicts with minimality of $n$. Now let $q$ be even, so that $a\mid 2^r+1$ and $b\mid 2^r+1$. Then, $a\mid 2^n+2^r=2^r(2^{n-r}+1)$, yielding $a\mid 2^{n-r}+1$. Likewise, $b\mid 2^{n-r}-1$. Now $n-r\ge 1$ and $n-r<k$ contradicts with the minimality of $k$.
Case 2: $k<n$. This is analogous to above, and omitted.