Infinite sum involving harmonic numbers

Question

How can I find a closed form formula for the following sum?
For $p\in\mathbb{Z}^+\land q\in\mathbb{Z}^+\setminus\{1\}$,
\begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^p}{k^q} \end{equation*} Only if $p\leqslant2$ and $q\leqslant2$, this sum is following: \begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^p}{k^q}=\dfrac{(pq-q\delta_{p,1})^2+1}{p+q-q\delta_{p,1}}\zeta(p+q) \end{equation*} where $H_k$ is following \begin{equation*} H_k=\displaystyle\sum_{j=1}^{k}\dfrac{1}{j} \end{equation*} When $p=1$ and $q=2$,
\begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^1}{k^2}=2\zeta(3) \end{equation*} When $p=2$ and $q=2$,
\begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^2}{k^2}=\dfrac{17}{4}\zeta(4) \end{equation*}

However, when ($p\geqslant3\land q\geqslant2$) or ($p\geqslant2\land p\geqslant3$), I've failed.
I think that my formula is wrong for all $p$ and $q$.
I've seen this answer.
Thank you.
p.s.
If $1\leqslant p\leqslant2$ and $2\leqslant q\leqslant4$, \begin{align*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^p}{k^q}=\dfrac{(pq-q\delta_{p,1})^2+1}{p+q-q\delta_{p,1}}\zeta(p+q) -\sum _{k=1}^{q-2} \left(\frac{1}{4} (-1)^{q+1} \binom{q+1}{k} \zeta (k+p) \zeta (q-k)+\frac{\left((p q)^2+3\right) \zeta (p+q)}{2 (p+q)}\right)-\frac{1}{2} q \left(\sum _{k=1}^{p-2} \zeta (p-k) \zeta (k+q)\right)-\sum _{k=1}^{q-3} \frac{(p q+5) \zeta (k+p)^2}{q} \end{align*} when $p=2$ and $q=3$, \begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^2}{k^3}=-\zeta(2)\zeta(3)+\dfrac{7}{2}\zeta(5) \end{equation*} when $p=2$ and $q=4$, \begin{equation*} \displaystyle\sum_{k=1}^\infty\dfrac{H_k^2}{k^4}=\dfrac{97\pi^6}{22680}-2\{\zeta(3)\}^2 \end{equation*}