Pascal-ish Triangle

Let $S_n=\dfrac{1}{1^n}+\dfrac{1+\frac{1}{2}}{2^n}+\dfrac{1+\frac{1}{2}+\frac{1}{3}}{3^n}+\cdots$.

Then, for positive even numbers $m$, there is a beautiful relationship between $S_{m}$ and the Riemann zeta function $\zeta(\cdot)$: $\begin{array} { l r c } S_2 &=& 2\zeta(3) \\ S_4 &=& {3\zeta(5)} \quad {-\zeta(2)\zeta(3)} \\ S_6 &=& {4\zeta(7)}\quad {-\zeta(2)\zeta(5)} \quad {-\zeta(3)\zeta(4)} \\ S_8 &=& {5\zeta(9)}\quad {-\zeta(2)\zeta(7)} \quad {-\zeta(3)\zeta(6)} \quad {-\zeta(4)\zeta(5)} \\ & \vdots & \\ S_{m} &=& \displaystyle \frac{m+2}2 \zeta(m+1) - \sum_{k=2}^{\frac m2} \zeta(k) \zeta(m-k). \end{array}$ However, there is also a relationship between positive odd numbers $m$ and the Riemann zeta function. Find this relationship and submit your answer as $\dfrac{\pi^4}{S_3}.$

Bonus: Prove the pattern shown above.