The solution of your differential equation with $Z(0) = Z_0$ is $ Z(t) = (e^t + (1-e^t) Z_0)^{-1} Z_0 $ as long as $e^t + (1-e^t) Z_0$ is invertible. In order for this to be periodic with period $p$, you'd need $(1-e^p) Z_0 (1-Z_0) = 0 $.

The solution $Z(t)$ of your differential equation with $Z(0) = Z_0$ satisfies $$ Z(t) (e^t + (1-e^t) Z_0) = Z_0 $$ In order for this to be periodic with period $p$, you'd need $(1-e^p) Z_0 (1-Z_0) = 0 $. $1-e^p = 0$ (for real $p$) only if $p=0$, while if $Z_0 (1-Z_0) = 0$ we have a fixed point.