Find all pairs $(m, n)$ of positive integers such that $m$ divides $8n+1$ and $n$ divides $8m+1$ I've found the pairs $(1,3),(1,9),(3,25)$ and $(13,21)$ up to order. But I have no idea how to prove that there are not other solutions. Any hints...? I've been trying for a few days but all I came up with is that seems that $8m+1$ or $8n+1$ is a square. 

Here is a solution based on the @abiessu's suggestion.
If $m\mid 8n+1$ and $n\mid 8m+1$, then $mn\mid (8m+1)(8n+1)$. In view of 
  $$ (8m+1)(8n+1) = 64mn+8m+8n+1, $$
this implies $mn\mid(8m+8n+1)$. As a result, $mn\le 8m+8n+1$; equivalently, $(m-8)(n-8)\le 65$. 
It follows that either $\max\{m,n\}\le 73$, or $\min\{m,n\}\le 8$. In either case we are left with a finite number of pairs to check; this is obvious in the first case where $\max\{m,n\}\le 73$, while in the first case, for each $m\le 8$ we must check only those $n$ dividing $8m+1$. Checking all these pairs, there are six solutions with $m\le n$ altogether:
  $$ (1,1),\ (1,3),\ (1,9),\ (3,25),\ (9,73),\ \text{and}\ (13,21). $$
