I recently found these Plouffe-like formulas using Mathematica's LatticeReduce. Has anybody seen/can prove these are indeed true?
$$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}$$
And so on for other $\zeta(2n+1)$. The background for these are in my blog.
In this paper by Vepstas he builds a derivation for Plouffe's identities, and generalizes them to some degree.
Your identities for $\zeta(4n-1)$ follow from the results immediately following his Corollary 2 $$ q^{k-1}\sum\frac{1}{n^k(e^{2\pi pn/q}-1)}+p^{k-1}\sum\frac{1}{n^k(e^{2\pi qn/p}-1)} = q^{k-1}I_k(2\pi p/q) $$ with $p=\sqrt{2},q=1$, and using the expression for $I_k(x)$ from the middle of page 7 (I confirmed for $k=3$ and expect it will also match for $k=7$).
He uses additional machinery to establish the identities in Plouffe's form for $k=4n+1$, and it doesn't immediately admit generalization to your form, but maybe it can be tweaked.