Does frequency at zero matter?
The reason why you do not "see" the zero is because it lies at the origin, for \$s=0\$. You can see that when \$s=0\$, the numerator is 0 and so is the magnitude: the dc component is blocked by the capacitor.

There is actually a better way to write this transfer function because it does not provide insight on the high-frequency plateau. This form is called a low-entropy form, a term forged by the late Dr. Middlebrook some time ago. It implies that a transfer function linking a response (\$V_{out}\$) to a stimulus (\$V_{in}\$) must be preferably expressed in way where zeroes, poles and gains properly appear. In your case, you can advantageously factor \$sR_2C_1\$ in the numerator and \$s(R_1+R_2)C_1\$ in the denominator:

\$H(s)=\frac{sR_2C_1}{s(R_1+R_2)C_1}\frac{1}{1+\frac{1}{s(R_1+R_2)C_1}}=H_{\infty}\frac{1}{1+\frac{\omega_p}{s}}\$ in which:

\$H_{\infty}=\frac{R_2}{R_1+R_2}\$ and \$\omega_p=\frac{1}{C_1(R_1+R_2)}\$

In this expression, \$\omega_p\$ is an inverted pole and lets you shape the transfer function in this convenient way, with a leading term \$H_{\infty}\$ representing the plateau gain as \$s\$ approaches infinity.

This is the proper way of writing this transfer function. If you are interested by determining transfer functions in a quick and swift way, you can have a look at the fast analytical circuits techniques or FACTs for which an introduction is given here.