The pole of a 1st-order system is the inverse of its natural time constant obtained when the excitation is removed. Here, if you temporarily remove the cap. and "look" through its terminals while \$V_{in}\$ is zeroed, then because of the virtual ground, the time constant is zero.
So \$D(s)=1+s\tau=1\$.
Regarding the numerator, if you bring the stimulus back in place, what condition would prevent the stimulus from generating an ac signal on the output? In other words, what condition nulls the output? Look at the below circuit, it is when the impedance made of \$C_1\$ paralleled with \$R_1\$ approaches infinity:
It becomes infinite at the pole frequency of this impedance. The time constant is \$R_1C_1\$. Therefore, \$N(s)=1+s\tau_N=1+\frac{s}{\omega_z}\$ with \$\omega_z=\frac{1}{R_1C_1}\$.
The final transfer function is thus defined as \$H(s)=H_0(1+s\tau_N)=H_0(1+\frac{s}{\omega_z})\$ with \$H_0=-\frac{R_2}{R_1}\$
In my opinion, describing transfer functions via the time constants concept represents the best way as you can later infer or sense where the poles and zeroes are located when inspecting an electrical diagram whether it is passive or active. I have described the fast analytical techniques or FACTs in a book I published in 2016 but you can have a look at the seminar I taught at APEC in 2016, it is a quick introduction to the subject. If your teacher has the FACTs in mind, he must be encouraged to pursue this path!
