How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$

You're almost there.

$$\frac{1}{m^2 + (N-m)^2} = \frac{1}{2m^2 +N^2 -2mN} = \frac{1}{2m(m-N) +N^2}\ge \frac{1}{N^2}$$

Now $$\sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{m^2 + (N-m)^2} \ge \sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{N^2} = \sum_{N=1}^{\infty}\frac{N-1}{N^2}$$

Can you finish from here?

If the sum were finite, then we could get a contradiction as follows. Breaking it up into 4 sums depending on whether or not $m$ and $n$ are even, we have

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2} = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n)^2} $$ $$+ \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n-1)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n-1)^2} $$ Note that each of the last three sums is greater than the first due to the denominators of each term being smaller. Thus we have $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2} > 4 \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} $$ But factoring out the 4 from the denominator we see the right hand side is the same as the left. Hence we get a contradiction and the sum is not finite.