How to determine poles and zeros of the z-transform? This is a simple question, but I just don't understand how we determine the poles and zeros of a rational system function.

For example, for the LTI system described by this constant coefficient difference equation
$$
y[n]-\frac{5}{2}y[n-1]+y[n-2]=x[n]
$$
we can determine that

$$
H(z)=\frac{1}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}
$$

I understand why there are poles at $z=\frac{1}{2}$ and $z=2$, but I don't understand why there are two zeros at $z=0$. Even if I multiply $H(z)$ through by $z$, there would only be one zero, correct?
Just multiply numerator and denominator by $z^2$ to obtain

$$H(z)=\frac{z^2}{(z-\frac12)(z-2)}$$

from which you see that there's a double zero at $z=0$. Note that the number of zeros and poles is always equal if you include poles and zeros at infinity.