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?

We have
$$
\sum^{\infty}_{n=1}\frac{2^n-1}{2^n}\frac{x^n}{n^2}=\frac{x}{(1-x)(2-x)}\textrm{, }|x|<1.
$$
Hence
$$
I=\int^1_0\int^1_0\frac{xy}{(1-xy)(2-xy)}dxdy=\int^{1}_{0}\int^{1}_{0}\sum^{\infty}_{n=1}\left(1-2^{-n}\right)(xy)^ndxdy=
$$
$$
=\sum^{\infty}_{n=1}\left(1-2^{-n}\right)\left(\int^{1}_{0}x^ndx\right)^2=\sum^{\infty}_{n=0}\frac{1}{(n+1)^2}-2\sum^{\infty}_{n=0}\frac{1}{2^{n+1}(n+1)^2}.
$$
Hence using $\sum^{\infty}_{n=0}\frac{x^{n}}{n+1}=-\frac{\log(1-x)}{x}$, $|x|<1$, we get
$$
I=-2\int^{1}_{1/2}\frac{\log(1-x)}{x}dx+\int^{1}_{0}\frac{\log(1-x)}{x}dx=
$$
$$
=-\int^{1}_{1/2}\frac{\log(1-x)}{x}dx+\int^{1/2}_{0}\frac{\log(1-x)}{x}dx=
$$
$$
=-\int^{1/2}_{0}\frac{\log(x)}{1-x}dx
+\int^{1/2}_{0}\frac{\log(1-x)}{x}dx=\int^{1/2}_{0}\frac{d}{dx}\left[\log(1-x)\log(x)\right]dx=\log^22
$$
