Proving that $\sum_{(m,n)\in \Bbb Z \times \Bbb Z}\frac{1}{m^2+n^2+1}$ diverges.

You can compare with an integral, as in the case of the harmonic series when you compute the integral of $x^{-1}$.

Note that

$$\int_{\Bbb R^2}{dxdy\over x^2+y^2+1}=\int_{0}^{2\pi}\int_0^\infty {rdrd\theta\over r^2+1}=\infty.$$

and by sampling the function over boxed sectors $[n,n+1)\times [m,m+1)$ you have that the upper and lower sums bound the integral on each side and are equal (up to removal of finitely many terms) your sum.

It includes $\frac{1}{p+1}$ for every prime $p\equiv 1\pmod 4$, and $\sum_{p\equiv 1\pmod 4} \frac{1}{p}$ diverges by Dirichlet. Then use $\frac{1}{p+1}>\frac{1}{2p}$.

By the Gauss circle problem, it is well known that the number of lattice points in the set $\{x^2+y^2\leq R^2\}$ is given by $\pi R^2 + O(R)$, hence the series is divergent by comparison with the harmonic series, once we estimate the number of lattice points in the annulus $R_1^2\leq x^2+y^2\leq R_2^2$.