This is (almost) rehashing what has been said but a with an intent to spell things out.
Begin with the observation that $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\frac14\sum_{k=1}^{\infty}\frac{r_2(k)}{k^s}-\zeta(2s).$$ Using $r_2(k)=4\sum_{d\vert k}\left(\frac{-4}k\right)=4(1*\left(\frac{-4}k\right))(k)$, where $\left(\frac{a}b\right)$ is the Jacobi symbol, and $\left(\frac{-4}{2k}\right)=0, \left(\frac{-4}{2k+1}\right)=(-1)^k$, while the Dirichlet series evolves under the arithmetic convolution, i.e., $\sum\frac{(a*b)(k)}{k^s}=\sum\frac{a(k)}{k^s}\sum\frac{b(k)}{k^s}$, it follows that \begin{align}\sum_{k=1}^{\infty}\frac{r_2(k)}{k^s} &=4\sum_{k=1}^{\infty}\frac{(1*\left(\frac{-4}k\right))(k)}{k^s} =4\sum_{n=1}^{\infty}\frac1{n^s}\sum_{k=1}^{\infty}\frac{\left(\frac{-4}k\right)(k)}{k^s} \\ &=4\zeta(s)\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}=4\zeta(s)\beta(s). \end{align}