Amazingly, we could find rather good approximations of the second solution approximating $f'(x)$ by a $[2,n]$ Padé approximant built at $x=2$ and get the appoximate solution at the price of a linear equation.
The second degree in numerator is required since $x=2$ is a known solution. So, we shall have, as an approximation
$$f'_{(n)}(x)=(x-2)\frac {a^{(n)}_0+a^{(n)}_1(x-2) } {1+\sum_{k=1}^n b_k (x-2)^k}$$ in which all coefficients are expressed in terms of higher derivatives. Moreover, for any $n$, we shall have $a^{(n)}_0=8 \log(2)$ and, then
$$x_{(n)}=2-\frac{a^{(n)}_0 } {a^{(n)}_1 }=2-\frac{8 \log(2) } {a^{(n)}_1 }$$ Moreover $a^{(n)}_1$ is itself a polynomial in $\log(2)$ with no constant term making the expression
$$x_{(n)}=2-\frac{8 } {\frac{a^{(n)}_1}{\log(2)} }=2-\frac 8 {c_{n}}$$
The first terms would be
$$\left(
\begin{array}{cc}
n & c_n \\
0 & {6 \left(2 +\log (2)\right)} \\
1 & \frac{2 \left(108 +60 \log (2)+11 \log ^2(2)\right)}{9
(2+\log (2))} \\
2 & \frac{2 \left(648 +396 \log (2)+86 \log ^2(2)+5 \log
^3(2)\right)}{ 108+60 \log (2)+11 \log ^2(2)} \\
3 & \frac{2 \left(58320 +38880 \log (2)+9720 \log ^2(2)+840 \log
^3(2)-43 \log ^4(2)\right)}{15 \left(648+396 \log (2)+86 \log ^2(2)+5
\log ^3(2)\right)}
\end{array}
\right)$$ and the decimal representation of the successive $x_{(n)}$ would then be
$$\left(
\begin{array}{cc}
n & x_{(n)} \\
0 & 1.50492 \\
1 & 1.37399 \\
2 & 1.35835 \\
3 & 1.35785
\end{array}
\right)$$
