Summing inverse squares of primitive Pythagorean triples

Part of the solution was already answered here:Sum of the inverse squares of the hypotenuse of Pythagorean triangles

As we all know, a primitive Pythagorean triple is determined by $(m^2-n^2,2mn,m^2+n^2)$, where $m$ and $n$ satisfy the following conditions:\begin{equation*}K = \{(m,n)\in \mathbb{N}: (m>n>0) \land (m\neq n \mod 2) \land (\mathrm{gcd}(m,n) = 1) \}\end{equation*}

From this, we can obviously conclude that the hypothenuse $H$ will always be of the form $m^2+n^2$, yeilding\begin{equation*}\sum \frac{1}{H^2} = \sum_K \frac{1}{(m^2+n^2)^2} = \frac{\mathrm{G}}{\zeta(2)}-\frac{1}{2}\end{equation*}Here, $\mathrm{G}$ stands for Catalan's constant, $\beta(2)$; Dirichlet's beta function evaluated at $s=2$. In general, for each quadratic form $q(m,n) = am^2+bmn+cn^2$, if the series $S^+(a,b,c;s)$ and $S^+_{12}(a,b,c;s)$ converge, the sum\begin{equation}\sum_K q(m,n)^{-s} = \frac{S^+(a,b,c;s)-S^+_{12}(a,b,c;s)}{4(1-2^{-2s})\zeta(2s)}\end{equation}Where $S^+ = \sum_{\mathbb{N}^2} q(m,n)^{-s}$ and $S^+_{12} = \sum_{\mathbb{N}^2} (-1)^{m+n} q(m,n)^{-s}$. However, a huge problem arises when you consider the small and large legs. That is, sometimes $m^2-n^2$ will be the small leg and vice versa. Solving the inequality

\begin{equation*}m^2-n^2 < 2mn \Longrightarrow m \in (n,n+n\sqrt{2}) \cap \mathbb{N} \cap \pi_1(K) = \{n+j:j\in \Gamma_n\}\end{equation*}Where, for each $n$, $\Gamma_n$ is the set of conditions\begin{equation*}\Gamma_n = \{ j\in\mathbb{N}: (0<j\leq \lfloor n\sqrt{2} \rfloor) \land (j\neq 0 \mod 2) \land (\mathrm{gcd}(j,n) = 1) \}\end{equation*}With this in mind, I have reached the following equalities:\begin{equation*}\sum \frac{1}{c_1^2} = \frac{1}{8} - M\end{equation*}\begin{equation*}\sum \frac{1}{c_2^2} = \frac{S^+(1,0,-1;2)\Big|_{m\neq n} - S_{12}^+(1,0,-1;2)\Big|_{m\neq n}}{4(1-2^{-4})\zeta(4)} + M\end{equation*}Where\begin{equation*}M = \sum_{n = 1}^{\infty}\sum_{\Gamma_n} \left( \frac{1}{4(n+j)^2n^2} - \frac{1}{(2nj+j^2)^2} \right) \end{equation*}Now my problem is how to compute $M$, $S^+(1,0,-1;2)\Big|_{m\neq n}$ and $S^+_{12}(1,0,-1;2)\Big|_{m\neq n}$, in terms of known constants if possible. Any ideas?

(PS: I also have a couple of cool problems regarding this one about interpolation and geometry, but first I need results for this very problem, specially to compute the sum of the $1/c_2^2$. I don't care as much about $1/c_1^2$, but obviously if we get a solution of one, we get the solution of the other.)

UPDATE. After careful examination, and with the great help of Mariusz Iwaniuk, we have an answer for the sum $\sum_K (m^2-n^2)^{-2}$.

$$\sum \frac{1}{c_2^2} = M + \frac{4}{15\zeta(4)} \sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty} \frac{2}{(m^2-n^2)^2} = M + \frac{\pi^4/96}{15/4 \cdot \pi^4/90} = \frac{1}{4} + M$$

The discussion on the computation of $M$ has been redirected to this other question:

How to sum $ \displaystyle\sum_{n=1}^\infty \sum_{\substack{j=1 \\\mathrm{gcd}(2n,j) = 1}}^{\lfloor n\sqrt{2}\rfloor} f(n,j) $