We can conclude that there exist $k, l \in \mathbb{N}$ such that $2m-1 = k n$ and $2n-1 = l m$. It follows that
$$4n - 2 = 2(2n - 1) = 2 \cdot l m = 2m \cdot l = (n k + 1) l.$$
From $4n - 2 = (n k + 1) l$ we get
$$(4 - k l) n = l + 2.$$
Since the right hand side is positive, it follows that $4 - k l > 0$, that is $k l < 4$. Since $k$ and $l$ are positive integers, we have the following possibilities:

a. $k l = 1$, that is $k = l = 1$. We get $2m - 1 = n$, $2n - 1 = m$, from which follows $m = n = 1$.

b. $k l = 2$. That is not possible, since from $2m - 1 = k n$, $2n - 1 = l m$ it follows that $k$ and $l$ are both odd.

c. $k l = 3$, that is $k = 3$, $l = 1$ or $k = 1$, $l = 3$. In the first case from $2m - 1 = 3n$, $2n - 1 = m$ we get $m = 5$, $n = 3$. Analogously, in the second case we get $m = 3$, $n = 5$.

Hence, all the solutions are $(m, n) \in \{(1, 1), (3, 5), (5, 3)\}$.