Conjecture $\sum_{m=1}^\infty\frac{y_{n+1,m}y_{n,k}}{[y_{n+1,m}-y_{n,k}]^3}\overset{?}=\frac{n+1}{8}$, where $y_{n,k}=(\text{BesselJZero[n,k]})^2$

There is a rather neat proof of this.

First, note that there is already an analogue for this: DLMF §10.21 says that a Rayleigh function $\sigma_n(\nu)$ is defined as a similar power series $$ \sigma_n(\nu) = \sum_{m\geq1} y_{\nu, m}^{-n}. $$ It links to http://arxiv.org/abs/math/9910128v1 among others as an example of how to evaluate such things.

In your case, call $\zeta_m = y_{\nu,m}$ and $z=y_{\nu-1,k}$ ($\nu$ is $n$ shifted by $1$), so that after expanding in partial fractions your sum is $$ \sum_{m\geq1} \frac{\zeta_m z}{(\zeta_m-z)^3} = \sum_{m\geq1} \frac{z^2}{(\zeta_m-z)^3} + \frac{z}{(\zeta_m-z)^2}. $$

Introduce the function $$ y_\nu(z) = z^{-\nu/2}J_\nu(z^{1/2}). $$ By DLMF 10.6.5 its derivative satisfies the two relations $$\begin{aligned} y'_\nu(z) &= (2z)^{-1} y_{\nu-1}(z) - \nu z^{-1} y_\nu(z) \\&= -\tfrac12 y_{\nu+1}(z). \end{aligned} $$

It also has the infinite product expansion $$ y_\nu(z) = \frac{1}{2^\nu\nu!}\prod_{k\geq1}(1 - z/\zeta_k). $$ Therefore, each partial sum of $(\zeta_k-z)^{-s}$, $s\geq1$ can be evaluated in terms of derivatives of $y_\nu$: $$ \sum_{k\geq1}(\zeta_k-z)^{-s} = \frac{-1}{(s-1)!}\frac{d^s}{dz^s}\log y_\nu(z). $$ When evaluating this logarithmic derivative, the derivative $y'_\nu$ can be expressed in terms of $y_{\nu-1}$, going down in $\nu$, but the derivative $y'_{\nu-1}$ can be expressed in terms of $y_\nu$ using the other relation that goes up in the index $\nu$. So even higher-order derivatives contain only $y_\nu$ and $y_{\nu-1}$.

I calculated your sum using this procedure with a CAS as: $$ -\tfrac12z^2(\log y)''' -z(\log y)'' = \tfrac18\nu + z^{-1} P\big(y_{\nu-1}(z)/y_\nu(z)\big), $$ where $P$ is the polynomial $$ P(q) = -\tfrac18 q^3 + (\tfrac38\nu-\tfrac18) q^2 + (-\tfrac14\nu^2 + \tfrac14\nu - \tfrac18)q. $$

When $z$ is chosen to be any root of $y_{\nu-1}$, $z=\mathsf{BesselJZero}[\nu-1, k]\hat{}2$, $P(q)=0$, your sum is equal to $$ \frac{\nu}{8}, $$ which is $(n+1)/8$ in your notation.

It is possible to derive a number of such closed forms for sums of this type. For example, by differentiating $\log y$ differently (going $\nu\to\nu+1\to\nu$), one would get $$ \sum_{m\geq1} \frac{y_{\nu,m}y_{\nu+1,k}}{(y_{\nu,m}-y_{\nu+1,k})^3} = -\frac{\nu}{8}. $$

Some other examples, for which the r.h.s. is independent of $z$ ($\zeta_m=y_{\nu,m}, z=y_{\nu-1,l}$, $l$ arbitrary): $$ \begin{gathered} \sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^2} = \frac14,\\ \sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^4} - \frac{1}{(\zeta_k-z)^2} + \frac1{24}\frac{5-\nu}{\zeta_k-z} = \frac{1}{48}, \\ \sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^4} + \frac1{96}\frac{z-\zeta_k-8+4\nu}{(\zeta_k-z)^2} = 0. \end{gathered} $$ or with $z=y_{\nu+1,l}$, $l$ arbitrary: $$ \begin{gathered} \sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^3} = -\tfrac18\nu-\tfrac14, \end{gathered} $$ and they get messier with higher degrees.

This is a partial answer concerning the last conjecture.

If no mistake : considering the partial fraction decomposition of $$A_m=\frac{m^2 x^2}{\left(m^2-x^2\right)^3}$$ $$A_m =\frac{m}{8 (m-x)^3}+\frac{m}{8 (m+x)^3}-\frac{1}{16 (m-x)^2}-\frac{1}{16 (m+x)^2}-$$ $$\frac{1}{16 m (m-x)}-\frac{1}{16 m (m+x)}$$ and then $$S(x)=\sum_{m=1}^\infty A_m=\frac \pi{16x}\Big(\cot (\pi x)+\pi x \csc ^2(\pi x)-2 \pi ^2 x^2 \cot (\pi x) \csc ^2(\pi x)\Big)$$ Then, if $k$ is an integer, $$S(k-\frac 12)=\frac {\pi^2} {16}$$

You could be interested by this paper.

Not an answer yet. But the references cited for that Wikipedia page may let you get the answer.

Begin with formula $$ J_\alpha(z) = \frac{(z/2)^\alpha}{\Gamma(\alpha+1)} \prod_{n=1}^\infty\left(1-\frac{z^2}{j_{\alpha,n}^2}\right) $$ Source: Wikipedia . Here $j_{\alpha,n}$ is the $n$th positive zero of $J_\alpha$.

Take logarithmic derivative to get $$ \frac{\alpha}{z} - \frac{J_{\alpha+1}(z)}{J_\alpha(z)} = \frac{\alpha}{z} + \sum_{n=1}^\infty\frac{2z}{z^2-j_{\alpha,n}^2} $$ Replace $\alpha$ by $-\alpha$ and simplify $$ \frac{J_{\alpha-1}(z)}{J_\alpha(z)} = 2\sum_{n=1}^\infty\frac{z}{j_{\alpha,n}^2-z^2} $$ Plug in $z=j_{\alpha-1,k}$ a zero of $J_{\alpha-1}$: $$ 0 = 2 \sum_{n=1}^\infty \frac{j_{\alpha-1,k}}{j_{\alpha,n}^2-j_{\alpha-1,k}^2} \\ 0 = \sum_{n=1}^\infty \frac{1}{j_{\alpha,n}^2-j_{\alpha-1,k}^2} $$ We want something like this with the third power in the denominator.