I  Since z^n is bigger that y^n, it must follow that z exceeds y.
II Let x^n<=y^n; then it must follow that z^n<=2y^n
We are going to consider the ratio of y and z, with y=(z-1), the most favourable case (see Note).
III z^(z+n)/(z-1)^(z+n)=((z-1)/z)^(-n-z)
IV {((z-1)/z)^(-n-z)} lim(z->±infinity) ((-1+z)/z)^(-n-z) = e
V But e>2; hence there are no solutions if the power (z+n) exceeds z.

Note: Let, say, y=z/2; then z^(z+n)/(z/2)^(z+n)=2^(z+n)