Unfortunately you can't.

Remember, whenever you see the differential form of stochastic integral, it is actually not a differential. It is only an abbreviation of the integral. dw,dz has no mathematical meaning unless they appear with ∫.

So dz=zdw actually means ∫dz=∫zdw, if you move z down to the left it actually means ∫dz/z=∫dw.

These two things are surely not equivalent. Even under Riemann integral you can't do that, because z is a function of w and time t perhaps. So how can we solve it?

let y=ln(z), apply Ito formula to y leads to dy=1/z*dz-0.5*1/z^2*(dz)^2 (even this form is not formal because always remember it is an integral, and (dz)^2 is actually mathematically incorrect) but that doesn't bother us. You just take it as a mathematical sign and abbreviation.

so put dz=zdw into it gives: dy=dw-0.5*dt, take integral on both sides(from 0 to t) gives
y(t)-y(0)=w(t)-0.5t. substitute y with ln(z) gives ln(z(t))=ln(z(0))+w(t)-0.5t.

You know, brownian motion is not a "smooth" function. Besides the ordinary Riemann part, there is a part for quadratic variation(-0.5t).

The normal process of solving this kind of problem is to guess a proper function for z, like f(z) and then apply Ito formula.

Remember,in stochastic calculus whenever you move something for one side of the equation to the other, you must be very careful!
Hope help~