Let us write $m = d a$ and $n = d b$. Here $a$ and $b$ are relatively prime integers and $v = d a b$. Plugging this into the given equation we get $d a + 3 d b - 5 = 2 d a b - 11 d$, so

$$d a + 3 d b - 5 = 2 d a b - 11 d$$

which rearranges to

$$d a + 3 d b - 2 d a b + 11 d = 5$$

or

$$d(a + 3b - 2ab + 11) = 5$$

Since $d$ is a positive integer, $d$ must divide $5$. Thus, $d = 1$ or $d = 5$.

First, let us assume that $d = 1$. Then the equation can be rearranged into

$$a + 3b - 5 = 2ab - 11$$

or

$$a + 3b + 6 = 2ab - 11 + 6 + 5 = 2ab$$

so

$$a + 3b + 6 = 2ab$$

or

$$2ab - a - 3b - 6 = 0$$

or

$$a(2b - 1) = 3b + 6$$

From here we conclude that $2b - 1$ divides $3b + 6$, so it must also divide $2(3b + 6) - 3(2b - 1) = 6b + 12 - 6b + 3 = 15$.

Since $2b - 1$ is a positive integer there are four possibilities:

- If $2b - 1 = 1$, then $b = 1$ and $a = 9$.
- If $2b - 1 = 3$, then $b = 2$ and $a = 4$, which leads to a contradiction, since $a$ and $b$ should be relatively prime.
- If $2b - 1 = 5$, then $b = 3$ and $a = 3$, which is again a contradiction.
- If $2b - 1 = 15$, then $b = 8$ and $a = 2$, and once more this is not possible.

Hence, in this case we only have one solution, $(9, 1)$.

Now, assume that $d = 5$. Then we can divide the equation by $5$ and rearrange it to get

$$a + 3b - 1 = 2ab - 11$$

or

$$a + 3b + 10 = 2ab$$

so

$$a(2b - 1) = 3b + 10$$

This implies that $2b - 1$ divides $3b + 10$, so it must also divide $2(3b + 10) - 3(2b - 1) = 6b + 20 - 6b + 3 = 23$.

There are two possibilities:

- If $2b - 1 = 1$, then $b = 1$ and $a = 13$.
- If $2b - 1 = 23$, then $b = 12$ and $a = 2$, a contradiction once more.

In this case we get the solution $(65, 5)$.

Therefore, the solutions are $(m, n) = (9, 1)$ and $(65, 5)$.