jmc

Looking at the form of the equation, we see that we have two linear terms and their product. We thus apply Simon's Favorite Factoring Trick. The given equation rearranges to $6mn+3m+2n+1=28$, which can be factored to $(3m+1)(2n+1)=28=2\cdot 2\cdot 7$. Since $n$ is a positive integer, we see that $2n+1>1$ is odd. Examining the factors on the right side, we see we must have $2n+1=7$, implying $3m+1=2\cdot 2$. Solving, we find that $(m,n)=\boxed{(1,3)}$.