Solution:
Let us first consider the pairs $(m, n)$ with $m \leq n$. We have
$$\frac{2}{5}=\frac{1}{m}+\frac{1}{n}-\frac{1}{mn} \leq \frac{1}{m}+\frac{1}{m}-\frac{1}{mn}<\frac{2}{m}$$
from which $m<5$. Moreover
$$\frac{2}{5}=\frac{1}{m}+\frac{1}{n}-\frac{1}{mn}>\frac{1}{m}$$
from which $m>\frac{5}{2}$, that is, since $m$ is an integer, $m \geq 3$. Setting $m=3$ we obtain
$$\frac{1}{3}+\frac{1}{n}-\frac{1}{3n}=\frac{2}{5}$$
from which $n=10$. Setting $m=4$ we obtain
$$\frac{1}{4}+\frac{1}{n}-\frac{1}{4n}=\frac{2}{5}$$
from which $n=5$. Finally, considering the symmetric case in which $n \leq m$ we get that the solution pairs must belong to the set $\{(3,10),(4,5),(10,3),(5,4)\}$. It is finally immediate to verify that the four pairs above are indeed solutions of the given equation.

Second solution.
Multiplying the given equation by $10 mn$ (recall that $m$ and $n$ are nonzero), we obtain
$$4mn-10m-10n+10=0,$$
that is,
$$(2m-5)(2n-5)=15$$
Here too let us first assume that $m \leq n$. The only pairs of integers whose product is $15$, with the first term less than or equal to the second, are $(3,5),(1,15),(-5,-3)$ and $(-15,-1)$.
- Setting $2m-5=1,\ 2n-5=15$, we obtain $(m, n)=(3,10)$.
- Setting $2m-5=-5,\ 2n-5=-3$, we obtain $m=0$, which is not acceptable.
- Setting $2m-5=-15,\ 2n-5=-1$, we obtain $(m, n)=(-5,2)$, which is not acceptable since $m$ is negative.
Then, considering the pairs $(m, n)$ with $n<m$, we obtain that the solutions are $\{(3,10),(4,5),(10,3),(5,4)\}$.