In the 2023 preprint [arXiv:2312.14051], K. C. Au used the WZ method to confirm the identity $$\sum_{k=1}^\infty\frac{(21k^3-22k^2+8k-1)256^k}{k^7\binom{2k}k^7}=\frac{\pi^4}{8}\tag{1}$$ conjectured by J. Guillera in 2003.
For any positive integer $m$, the harmonic numbers of order $m$ are given by $$H_n^{(m)}:=\sum_{0<k\le n}\frac1{k^m}\ \ \ (n=0,1,2,\ldots).$$
Let $P(k):=21k^3-22k^2+8k-1$. Motivated by the identity $(1)$, I conjectured the identity $$\sum_{k=1}^\infty\frac{256^k}{k^7\binom{2k}k^7}\left(P(k)(4H_{2k-1}^{(2)}-5H_{k-1}^{(2)})-6k+2\right)=\frac{\pi^6}{24}\tag{2}$$ in 2022 (see Equation (178) in this paper), and also conjectured the identity $$\sum_{k=1}^\infty\frac{256^k}{k^7\binom{2k}k^7}\left(P(k)(16H_{2k-1}^{(4)}+7H_{k-1}^{(4)})+\frac 4k\right)=\frac{31\pi^8}{1440}\tag{3}$$ in 2023 (see Equation (5.20) in this paper).
In view of the identities $(2)$ and $(3)$, here I propose a further conjecture involving harmonic numbers of order six.
Conjecture. We have the formula $$\sum_{k=1}^\infty\frac{256^kP(k)(64H_{2k-1}^{(6)}-65H_{k-1}^{(6)})}{k^7\binom{2k}k^7}=\frac{31\pi^{10}}{3780}. \tag{4}$$ Also, for any prime $p\ge5$ we have the congruence $$\begin{aligned}&\sum_{k=0}^{p-1}\frac{\binom{2k}k^7}{256^k}(21k^3+22k^2+8k+1)\left(64H_{2k}^{(6)}-65H_k^{(6)}\right) \\&\qquad\ \ \equiv\frac{1488}5p^2B_{p-5}\pmod{p^3}, \end{aligned}\tag{5}$$ where $B_0,B_1,B_2,\ldots$ are the Bernoulli numbers.
As the series in $(4)$ has converging rate $1/64$, it is easy to check the identity $(4)$ numerically. I don't know whether there are some other series for $\pi^{10}$ with geometric converging rate. The $p$-adic congruence $(5)$ looks challenging.
QUESTION. Any ideas to prove the identities $(2)-(4)$? Can one modify Au's WZ-seed method to confirm $(4)$?
