For $m=1$, we get $n\mid 3$, and thus $n=3$. We get therefore $(1,3)$ and $(3,1)$ as two pairs. Now, note that both $m,n$ are odd; and assume $m,n\geqslant 3$. Suppose $m=\prod_{k=1}^N p_k^{\alpha_k}$ and $n=\prod_{k=1}^M q_k^{\beta_k}$. The idea is to look at the appropriate powers of two, as I outline below. Now, suppose $1\leqslant i \leqslant N$ and $1\leqslant j \leqslant M$ are such that $v_2(p_i-1) = \min_{1\leqslant k\leqslant N}v_2(p_k-1)$, and $v_2(q_j-1) = \min_{1\leq k\leq M}v_2(q_k-1)$, where $v_2(n) = \alpha \iff 2^{\alpha}\mid n$ and $2^{\alpha+1}\nmid n$. Now, observe that, $2^{\varphi(n)}\equiv -1\pmod{p_i^{\alpha_i}}$ and $2^{2\varphi(n)}\equiv 1\pmod{p_i^{\alpha_i}}$. In particular, if $d_i$ is the smallest positive integer for which $2^{d_i}\equiv 1\pmod{p_i^{\alpha_i}}$, we then have $d_i\nmid \varphi(n)$ and $d_i\mid 2\varphi(n)$, from which we get $v_2(d_i) =v_2(\varphi(n))+1$. Finally, since $d_i\mid p_i^{\alpha_i-1}(p_i-1)$ (by Euler's theorem), we also have $v_2(d_i)\leqslant v_2(p_i-1)$. Thus, $v_2(p_i-1)\geqslant 1+v_2(\varphi(n))\geqslant 1+v_2(q_j-1)$. Repeating the same logic, this time on the other congruence with $q_j^{\beta_j}$, we get immediately $v_2(q_j-1)\geqslant v_2(p_i-1)+1$. Adding these two up, we get $v_2(p_i-1)+v_2(q_j-1)\geqslant v_2(p_i-1)+v_2(q_j-1)+2$, which is a clear contradiction. Thus, no solutions in the regime $m,n\geqslant 3$.