$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket. Namely, $\ds{\bracks{P} = 1}$ whenever $\ds{P\ \mbox{is}\ \color{red}{\texttt{true}}}$ and $\ds{0\ \color{red}{\mbox{otherwise}}}$.
\begin{align}
\left.\sum_{n = 1}^{\infty}H_{n}^{2}x^{n}\,\right\vert_{\ \verts{x}\ <\ 1} & =
\sum_{n = 1}^{\infty}\overbrace{\braces{\sum_{i = 1}^{\infty}{\bracks{i \leq n} \over i}}}^{\ds{H_{n}}}\
\overbrace{\braces{\sum_{j = 1}^{\infty}{\bracks{j \leq n} \over j}}}^{\ds{H_{n}}}\ x^{n}
\\[5mm] & =
\sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j}
\sum_{n = 1}^{\infty}\bracks{n \geq i}\bracks{n \geq j}x^{n}
\\[5mm] & =
\sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j}\braces{%
\bracks{i \leq j}\sum_{n = j}^{\infty}x^{n} +
\bracks{i > j}\sum_{n = i}^{\infty}x^{n}}
\\[5mm] & =
\sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j}\braces{%
\bracks{j \geq i}\,{x^{j} \over 1 - x} +
\bracks{j < i}\,{x^{i} \over 1 - x}}
\\[5mm] & =
{1 \over 1 - x}\sum_{i = 1}^{\infty}{1 \over i}\pars{%
\sum_{j = i}^{\infty}{x^{j} \over j} + \sum_{j = 1}^{i - 1}{x^{i} \over j}} =
{1 \over 1 - x}\pars{\sum_{i = 1}^{\infty}{1 \over i}
\sum_{j = i}^{\infty}{x^{j} \over j} +
\sum_{i = 1}^{\infty}{x^{i} \over i}\overbrace{\sum_{j = 1}^{i - 1}{1 \over j}}
^{\ds{H_{i - 1}}}}
\\[5mm] & =
{1 \over 1 - x}\pars{\sum_{j = 1}^{\infty}{x^{j} \over j}\
\overbrace{\sum_{i = 1}^{j}{1 \over i}}^{\ds{H_{j}}}\ +
\sum_{i = 1}^{\infty}{x^{i + 1} \over i + 1}\,H_{i}} =
{1 \over 1 - x}\sum_{i = 1}^{\infty}
\pars{{x^{i} \over i} + {x^{i + 1} \over i + 1}}H_{i}
\\[5mm] & =
{1 \over 1 - x}\sum_{i = 1}^{\infty}H_{i}
\pars{x^{i}\int_{0}^{1}t^{i - 1}\,\dd t + x^{i + 1}\int_{0}^{1}t^{i}\,\dd t}
\\[5mm] & =
{1 \over 1 - x}
\pars{\int_{0}^{1}{1 \over t}\sum_{i = 1}^{\infty}H_{i}\pars{xt}^{i}\,\dd t + x\int_{0}^{1}\sum_{i = 1}^{\infty}H_{i}\pars{xt}^{i}\,\dd t}
\\[5mm] & =
{1 \over 1 - x}\int_{0}^{1}{1 + xt \over t}
\bracks{-\,{\ln\pars{1 - xt} \over 1 - xt}}\,\dd t =
-\,{1 \over 1 - x}\int_{0}^{x}{1 + t \over t\pars{1 - t}}\,
\ln\pars{1 - t}\,\dd t
\end{align}
because $\ds{\sum_{i = 1}^{\infty}H_{i}z^{i} =
-\,{\ln\pars{1 - z} \over 1 - z}:\ \pars{~H_{i}\ Generating\ Function~}}$.
Then,
\begin{align}
\left.\sum_{n = 1}^{\infty}H_{n}^{2}x^{n}\,\right\vert_{\ \verts{x}\ <\ 1} & =
-\,{1 \over 1 - x}\bracks{%
2\
\underbrace{\int_{0}^{x}{\ln\pars{1 - t} \over 1 - t}\,\dd t}
_{\ds{-\,{1 \over 2}\,\ln^{2}\pars{1 - x}}}\ +\
\underbrace{\int_{0}^{x}\overbrace{{\ln\pars{1 - t} \over t}}
^{\ds{-\,\mrm{Li}_{2}'\pars{x}}}\ \,\dd t}
_{\ds{-\,\mrm{Li}_{2}\pars{x}}}}
\\[5mm] & =
\bbx{\ln^{2}\pars{1 - x} + \mrm{Li}_{2}\pars{x} \over 1 - x}
\end{align}
