Why does Miller approximation method give the wrong result here?
Your mistake was in the original expression for the voltage gain (which you called \$A\$). The voltage gain in Miller's Theorem is the voltage gain of the circuit including the effects of the feedback impedance.
Miller's Theorem is valid when the output voltage is proportional to the input voltage with some constant of proportionality that I will call \$K\$ for consistency with the Wikipedia page on Miller's Theorem. We don't yet know what \$K\$ is, but we solve for it self-consistently using the re-drawn circuit with \$Z_1\$ and \$Z_2\$, which depend upon \$K\$. The small-signal voltage at the drain for a given change in the drain current is determined by \$Z_2\$ because of the (presumably) high output impedance of the current source and of \$r_o\$.
It is therefore easy to see that \$K = -g_mZ_2\$. According to Miller's Theorem, \$Z_2 = \frac{Z_f}{1-1/K}\$. Plug this expression for \$Z_2\$ into the expression for \$K\$ and solve for \$K\$:
$$K=-g_m\frac{Z_f}{1-1/K}$$
$$K(1-1/K) = -g_mZ_f$$
$$K = 1-g_mZ_f$$
The input impedance is determined entirely by \$Z_1\$, since the transistor has extremely high input impedance. Therefore \$Z_{eq} = Z_1\$, which according to Miller's Theorem, is \$\frac{Z_f}{1-K}\$. Thus,
$$Z_{eq} = \frac{Z_f}{1-K} = \frac{Z_f}{g_mZ_f} = 1/g_m$$