(This answers Question 2.)

Thanks to Cohen's 2022 paper, turns out there is a deg-$4$ and one can find polynomials $Q_k(n)$ for general deg-$k$ such that,

$$(n+1)^k s_{n+1} = Q_k (n)\, s_n - n^k s_{n-1}\qquad\tag{eq.1}$$

is a 3-term recurrence relation. This can then be used to create continued fractions with closed-forms. Recall the general cfrac for zeta $\zeta(k)$,

$$\zeta(k) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{2k}}{n^k+(n+1)^k}}} \qquad$$

We employ "denominators" with similar forms, the first being Apery's,

\begin{align} Q_3(n) &= n^3 + (n + 1)^3 + 4(2n + 1)^3\\ Q_4(n) &= n^4 + (n + 1)^4 + 2n^2 + 2(n + 1)^2\\ Q_5(n) &= n^5 + (n + 1)^5 + 6n^3 + 6(n + 1)^3\\ Q_6(n) &= n^6 + (n + 1)^6 + 3 n^4 + 3(n + 1)^4 -(2n + 1)^2 \end{align}

then we get the nice cfracs,

\begin{align} F_3 &= \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{6}}{Q_3(n)}}} = \frac{\zeta(3)}6\\[8pt] F_4 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{8}}{Q_4(n)}}} = \zeta(4)+4\,\zeta(2)-8 \\[8pt] F_5 &= \frac1{7 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{10}}{Q_5(n)}}} = \zeta(5)+3\,\zeta(3)-9/2\\[8pt] F_6 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{12}}{Q_6(n)}}} = \zeta(6)+4\,\zeta(4)+16\,\zeta(2)-32 \end{align}

and so on. Note: These converge slightly faster (with Apery's the fastest) because to the general $n^k + (n + 1)^k$ expression, more terms have been added.

P.S. For the recurrence, one may also choose exponents $\alpha,\beta,$

$$(n+1)^\alpha s_{n+1} = Q_k (n)\, s_n - n^\beta s_{n-1}\qquad$$

such that $\alpha \leq \beta$ and $\alpha+\beta = 2k.$ Eq.1 was just the case $\alpha=\beta=k.$

(This answers Question 2.)

Thanks to Cohen's 2022 paper, turns out there is a deg-$4$ and one can find polynomials $Q_k(n)$ for general deg-$k$ such that,

$$(n+1)^k s_{n+1} = Q_k (n)\, s_n - n^k s_{n-1}\qquad\tag{eq.1}$$

is a 3-term recurrence relation. This can then be used to create continued fractions with closed-forms. Recall the general cfrac for zeta $\zeta(k)$,

$$\zeta(k) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{2k}}{n^k+(n+1)^k}}} \qquad$$

We employ "denominators" with similar forms, including deg-$4$, the first being Apery's,

\begin{align} Q_3(n) &= n^3 + (n + 1)^3 + 4(2n + 1)^3\\ Q_4(n) &= n^4 + (n + 1)^4 + 2n^2 + 2(n + 1)^2\\ Q_5(n) &= n^5 + (n + 1)^5 + 6n^3 + 6(n + 1)^3\\ Q_6(n) &= n^6 + (n + 1)^6 + 3 n^4 + 3(n + 1)^4 -(2n + 1)^2 \end{align}

then we get the nice cfracs,

\begin{align} F_3 &= \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{6}}{Q_3(n)}}} = \frac{\zeta(3)}6\\[8pt] F_4 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{8}}{Q_4(n)}}} = \zeta(4)+4\,\zeta(2)-8 \\[8pt] F_5 &= \frac1{7 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{10}}{Q_5(n)}}} = \zeta(5)+3\,\zeta(3)-9/2\\[8pt] F_6 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{12}}{Q_6(n)}}} = \zeta(6)+4\,\zeta(4)+16\,\zeta(2)-32 \end{align}

and so on. Note: These converge slightly faster (with Apery's the fastest) because to the general $n^k + (n + 1)^k$ expression, more terms have been added.

P.S. For the recurrence, one may also choose exponents $\alpha,\beta,$

$$(n+1)^\alpha s_{n+1} = Q_k (n)\, s_n - n^\beta s_{n-1}\qquad$$

such that $\alpha \leq \beta$ and $\alpha+\beta = 2k.$ Eq.1 was just the case $\alpha=\beta=k.$