If $x^{{\left({\log }_{2}x\right)}^2-6{\log }_{2}x+1}=64$, then $x=$

$x^{{\left({\log }_{2}x\right)}^2-6{\log }_{2}x+1}=64$

${{\log }^{64}}_x={\left({{\log }_2}^x\right)}^2-6{{\log }^x}_2+1$

${\log }_x{2}^6={\left({{\log }_2}^x\right)}^2-6{{\log }_2}^x+1$

$6{{\log }_2}^x={\left({{\log }_2}^x\right)}^2-6{{\log }_2}^x+1$

$\frac{6}{{{\log }_2}^x}={\left({{\log }_2}^x\right)}^2-6{{\log }_2}^x+1$

${{\log }_2}^x=a$

$\frac{6}{a}=a^2-6a+1$

$a^3-6a^2+a-6=0$

$\left(x-6\right)\left(a^2+1\right)=0$

$a=6,a^2=-1$ (does not exist)

$a=6$

Replace $a={{\log }^x}_2\to {{\log }_2}^x=6$

$x=2^6,x=64$