Using the expansion
$$
\begin{align}
\frac1{z-x}
&=\frac1{z-1}\frac1{1+(1-x)/(z-1)}\\
&=\frac1{z-1}-\frac{1-x}{(z-1)^2}+\frac{(1-x)^2}{(z-1)^3}+\dots
\end{align}
$$
we get
$$
\begin{align}
\int_0^1\frac{\log^2(1-x)}{z-x}\,\mathrm{d}x
&=\int_0^1\frac{\log^2(1-x)}{z-x}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(z-1)^{k+1}}\int_0^1\log^2(1-x)(1-x)^k\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(z-1)^{k+1}}\int_0^1\log^2(x)x^k\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(z-1)^{k+1}}\int_0^\infty t^2e^{-(k+1)t}\,\mathrm{d}t\\

@anon We prove that $\lim_{n\to\infty}\sum_{k=n+1}^\infty x_n^2=0$ if $\sum_{k=1}^\infty x_k^2<\infty$ as follows, right?

For each $n,m$ with $m>n$ we have:

$\sum_{k=n+1}^m x_k^2= \sum_{k=1}^m x_k^2- \sum_{k=1}^n x_k^2$

We fix a $n$ and take the limit as $m \to +\infty$.

$\sum_{k=n+1}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^n x_k^2$

Now we let $n \to +\infty$ and we have:

$\lim_{n \to +\infty}\sum_{k=n+1}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^{+\infty} x_k^2=0$