A double sum for the square of the natural logarithm of $2$. I am trying to show

\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^2 2^{n}} =(\ln(2))^2. 
\end{eqnarray*}

Motivation : I want to use this to calculate $ \operatorname{Li}_2(1/2)$. So I want a solution to the above that does not use any reference to dilogarithms and please avoid rational multiples of $\pi^2$ (if possible).
Right, lets turn it into a double integral. ( I know lots of you folks prefer integrals to plums.)
Show

\begin{eqnarray*}
\int_0^1 \int_0^1 \frac{xy \ dx  \ dy}{(1-xy)(2-xy)} =(\ln(2))^2. 
\end{eqnarray*}

Reassuringly Wolfy agrees
My try: Let $u=xy$, & the double integral becomes
\begin{eqnarray*}
\int_0^1 \frac{dy}{y}  \int_0^y \frac{u \ du }{(1-u)(2-u)} . 
\end{eqnarray*}
Partial fractions
\begin{eqnarray*}
 \frac{u}{(1-u)(2-u)} =\frac{1}{1-u} - \frac{2}{2-u}.
\end{eqnarray*}
Do the $u$ integrations to leave the $y$ integrals
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  +2 \int_0^1 \frac{\ln(2-y) dy}{y}.
\end{eqnarray*}
The first integral is
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  = \frac{ \pi^2}{6}.
\end{eqnarray*}
which I was hoping to avoid and even worse Wolfy says the second integral is divergent
So you have a choice of questions, where did I go wrong in the above ? OR how can we show the initially stated result?

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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\ds{\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over \pars{n + m}^{2}\, 2^{n}} = \ln^{2}\pars{2}} \approx 0.4805:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over \pars{n + m}^{2}\, 2^{n}}} =
\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over 2^{n}}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{m + n - 1}\,\dd x}}
^{\ds{{1 \over \pars{n + m}^{2}}}}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
\sum_{n = 1}^{\infty}\pars{x \over 2}^{n}
\sum_{m = 1}^{\infty}x^{m}\,{\dd x \over x} =
-\int_{0}^{1}\ln\pars{x}
{x/2 \over 1 - x/2}\,{x \over 1 - x}\,{\dd x \over x}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
{x \over \pars{2 - x}\pars{1 - x}}\,\dd x =
2\int_{0}^{1}
{\ln\pars{x} \over 2 - x}\,\dd x - \int_{0}^{1}
{\ln\pars{x} \over 1 - x}\,\dd x
\\[5mm] = &\
2\int_{0}^{1/2}
{\ln\pars{2x} \over 1 - x}\,\dd x -
\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
-2\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\,\dd x +
\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\,\dd x =
-2\,\mrm{Li}_{2}\pars{1 \over 2} + \mrm{Li}_{2}\pars{1}
\\[5mm] = &\
-2\
\underbrace{\bracks{{\pi^{2} \over 12} -
{1 \over 2}\ln^{2}\pars{2}}}_{\ds{\mrm{Li}_{2}\pars{1 \over 2}}}\ +\
\underbrace{\pi^{2} \over 6}
_{\ds{\mrm{Li}_{2}\pars{1}}} = \bbx{\ln^{2}\pars{2}} \\ &
\end{align}
