Why does the high pass filter have a zero but the low pass doesn't?

The low-pass filter transfer function is $\hat{H} (\hat{s}) =\frac{1}{1+j\omega RC}$, which is never zero for any value of $\omega$. If you write it as a quotient of factorized polynomials, the denominator is just 1, which has no roots. In contrast, the high-pass filter transfer function is $\hat{H} (\hat{s}) =\frac{j \omega RC}{1+j\omega RC}$. The numerator is zero when $\omega=0$ (only), so it has one zero.