On generalizing the harmonic sum $\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n = S_{k-1,2}(1)+\zeta(k+1)$ when $z=1$?

After my last edit, I figured out a way to partially answer my question. The trick is to test,

$$F_k(z) - G_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n - \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^k}z^{n+1}$$

for various values of $k,z$ to see if it yields something familiar. For $k=2$ and value $z = 1/3$, the Inverse Symbolic Calculator was able to recognize it as,

$$F_2\big(\tfrac13\big) - G_2\big(\tfrac13\big) = \rm{Li}_3\big(\tfrac13\big)$$

A little more testing showed that for $-1\leq z\leq 1$, apparently,

$$F_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + S_{k,1}(z)$$

with Nielsen generalized polylogarithm $S_{n,p}(z)$. Equivalently, in terms of polylogarithm $\rm{Li}_n(z)$,

$$F_k(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{k+1}(z)$$

For the special cases $z=1$ and $z=-1$, the polylogarithm reduces to formulas $(2)$ and $(3)$ in the post.

P.S. Of course, what remains is to rigorously prove the proposed formula.