How does one calculate a pole-zero plot?
Suppose you are given a system with transfer function

$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})} $$

Poles

Poles are the values of $z$ for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of $z$ will turn the transfer function tend to infinity? Obviously it's $z= 4$ and $z=6$, because if you let $z$ equal 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

Zeros

Zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are $z= 3$ and $z=7$, cause if you put $z= 3$ or $z=7$, the numerator will be zero, that means the whole transfer function will be zero. 

So here poles are $z=4$ and $z=6$, and zeros are $z=3$ and $z=7$.

Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system:



So, for low pass filter, you find out the transfer function, then the poles and zeros.