$\sqrt[n]{(x+1)^2}+\sqrt[n]{(x-1)^2}=4\sqrt[n]{x^2-1}\Big/:2$
$\frac{\sqrt[n]{(x+1)^2}}{2}-\sqrt[n]{x^2-1}+\frac{\sqrt[n]{(x-1)^2}}{2}=\sqrt[n]{x^2-1}$
$\Bigg(\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}\Bigg)^{2}-2\cdot\Bigg(\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}\Bigg)+\Bigg(\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}\Bigg)^{2}=\sqrt[n]{x^2-1}$
$\Bigg(\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}-\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}\Bigg)^{2}=\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}-\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}$
$\Bigg|\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}-\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}\Bigg|=\sqrt[2n]{x^2-1}$
$\frac{\sqrt[n]{(x+1)}}{\sqrt{2}}-\frac{\sqrt[n]{(x-1)}}{\sqrt{2}}=\sqrt[2n]{x^2-1}$
I don't know how to continue.
Task from a contest in Poland, 1960
Let $(x+1)^{1/n}=a, (x-1)^{1/n}=b$
$\implies a^2+b^2=4ab\iff\left(\dfrac ab\right)^2-4\cdot\dfrac ab+1=0$
$\dfrac ab=2\pm\sqrt3$
$$\implies\dfrac{x+1}{x-1}=\left(\dfrac ab\right)^n$$