Consider that you look for the zero's of function
$$f(x)=2^{2x-3}-32(x-1)$$
By inspection, $f(0)=\frac{257}{8}$, $f(1)=\frac 12$, $f(2)=-30$; so, a solution close to $x=1$.
Perform a Taylor expansion around $x=1$ (this is equivalent to the first interation of Newton method); it will give
$$f(x)=\frac{1}{2}+(x-1) (\log (2)-32)+O\left((x-1)^2\right)$$ Ignoring the higher order terms, this would give, as an estimate,
$$x=\frac{2 \log (2)-65}{2 (\log (2)-32)}\approx 1.015970944$$
We could do better building a $[1,n]$ Padé approximant; in such a way, the estimate is obtained solving a linear equation in $(x-1)$. For example, the simplest (corresponding to the first iteration of Halley method) would be
$$f(x)=\frac{\frac{1}{2}+\frac{ \left(2048+\log ^2(2)-128 \log (2)\right)}{2 (\log
(2)-32)}(x-1)}{1-\frac{ \log ^2(2)}{\log (2)-32}(x-1)}$$ giving
$$x=\frac{2080+\log ^2(2)-129 \log (2)}{2048+\log ^2(2)-128 \log (2)}\approx 1.015974860$$ We could continue the process and get the following results
$$\left(
\begin{array}{cc}
n & x_{(n)} \\
0 & 1.0159709442173479658 \\
1 & 1.0159748596445568773 \\
2 & 1.0159748895153359604 \\
3 & 1.0159748896898660072 \\
4 & 1.0159748896907116852 \\
5 & 1.0159748896907153030 \\
6 & 1.0159748896907153175
\end{array}
\right)$$
Edit
If we are concerned by the roots, consider that
$$f'(x)=2^{2 x-2} \log (2)-32 \qquad \text{and} \qquad f''(x)=2^{2 x-1} \log ^2(2) \,\, > 0 \,\,\forall x$$ The first derivative cancels at
$$x_*=1+\frac{\log \left(\frac{32}{\log (2)}\right)}{2 \log (2)}\implies f(x_*)=\frac{16 \left(1+\log \left(\frac{\log (2)}{32}\right)\right)}{\log (2)} <0$$ So, two roots. To approximate them, build a Taylor series around $x_*$ to get
$$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O((x-x_*)^2)$$ and solve the quadratic; this will give $x_1=2.05$ and $x_2=5.48$. Using thse guesses, Newton method should converge quite fast.