This is not an answer, but just a sketch of a possible answer; I think the answer is no. For $T$ as desired, consider $Det(I-z^2T^2)=Det(I-zT)*Det(I+zT)\equiv 1$. Considering the Taylor expansion of this function, one can conclude that $Trace(T^{2n})=0$ for any natural $n$. I think this yields $T^2=0$, and then the claim follows.

Actually, I am not sure if such determinant may have positive exponential type - this could be another approach.

This is not an answer, but just a sketch of a possible answer; I think the answer is no. For $T$ as desired, consider $Det(I-z^2T^2)=Det(I-zT)\cdot Det(I+zT)=e^{-z}\cdot e^z\equiv 1$. Considering the Taylor expansion of this function, one can conclude that $Trace(T^{2n})=0$ for any natural $n$. I think this yields $T^2=0$, and then the claim follows.

Actually, I am not sure if such determinant may have positive exponential type - this could be another approach.