Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$
Long Comment:
The sum can be written in terms of unsigned Stirling numbers of the first kind, $s_{n}^{(k)}=\left[_{k}^{n} \right]$
$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n+1}^{(2)} x^n}{n!\; n^{\alpha}}$$
Since $H_n=\left[_{\,\,\,2}^{n+1} \right]\frac{1}{n!}$
The well known recurrence relation for unsigned Stirling numbers of the first kind is
$$\left[_{\,\,\,k}^{n+1} \right]=n\left[_{k}^{n} \right] +\left[_{k-1}^{\,\,n} \right]$$
which immediately leads to
$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n}^{(2)} x^n}{n!\; n^{{\alpha}-1}}+\sum _{n=1}^{\infty } \frac{ x^n}{ n^{{\alpha}+1}}\tag{1}$$
which can (according to Mathematica) be rewritten simply as (with $x=1$)
$$\sum_{n=1}^{\infty}\frac{H_n}{n^{\alpha}}=S_{{\alpha}-1,2}(1)+\text{Li}_{{\alpha}+1}(1)=S_{{\alpha}-1,2}(1)+\zeta({\alpha}+1)$$ where $S_{n,p}(z)=\frac{(-1)^{n+p-1}}{(n-1)! p!}\int _0^1\frac{ t \log ^{n-1} \left(\log ^p (1-t z)\right)}{t} \,dt$ is the Nielsen generalized polylogarithm function and $\text{Li}_n$ is the polylogarithm function.
[Update: The full infinite series form for the Nielsen generalized polylogarithm according to Mathematica is
$$S_{n,p}(z)=\sum _{k=1}^{\infty } \frac{ s_k^{(p)}}{k! \,k^n}z^k$$
However I cannot find this result online.]
The Nielsen generalized polylogarithm function is utilised in the study of quantum electrodynamics according to information available online, so its properties have been quite well studied apparently. I didn't have time to look further, but hopefully this gives one potential starting point in your quest for an answer.
Euler proved the following result:
Theorem For integer $q\geq 2$, we have $$\sum_{n=1}^\infty \frac{H_n}{n^q}=\left( 1+\frac{q}{2}\right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1) \zeta(q-k)$$
A proof using the residue theorem can be found in the paper: "Euler sums and contour integral representations" by Philippe Flajolet and Bruno Salvy.