How can we show that $y_n=z_n$? 
Let $$y_{n+1}=2^n\sqrt{2(1-\sqrt{1-(\frac{y_n}{2^n})^2}})$$ and $$z_{n+1}=\frac{2z_n}{\sqrt{2(1+ \sqrt{1- (\frac{z_n}{2^n})^2}})}$$

$y_1=z_1=2$
How can one shows that $z_n=y_n$?

I have assumed that $y_n=z_n$ then if that implies $y_{n+1}=z_{n+1}$ then by induction they are equal.  But I feel this is not a way to show that $y_n=z_n$

see that, $$1-\sqrt{1-(\frac{y_n}{2^n})^2} = \frac{\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)\left(1+\sqrt{1-(\frac{y_n}{2^n})^2}\right)}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} = \frac{(\frac{y_n}{2^n})^2}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} $$
Hence $$y_{n+1}=2^n\sqrt{2\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)}=2^n\sqrt{\frac{\color{blue}4(\frac{y_n}{2^n})^2}{\color{blue}2(1+\sqrt{1-(\frac{y_n}{2^n})^2})}} =\frac{2y_n}{\sqrt{2(1+ \sqrt{1- (\frac{y_n}{2^n})^2}})}$$
The result easily follows since $y_1=z_1$
