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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.