Alternatively, you can solve the ODE and recover the result:
With regards to the SDE for $X_t$, I got the same ODE as described above. We can derive the solution, $m(t)$, to the ODE as follows:
\begin{align}
\frac{dm(t)}{dt} &= \frac{1}{2} (iz)^2\cdot m(t)\\
&\Updownarrow\\
\frac{2\cdot\frac{dm(t)}{dt}}{m(t)} &= -z^2\\
&\Updownarrow\\
\int \frac{2\cdot\frac{dm(t)}{dt}}{m(t)} \: dt&= \int-z^2\: dt \\
&\Updownarrow\\
2\ln\left(m(t)\right) &= -z^2t+C_1\\
&\Updownarrow\\
m(t)&=e^{\frac{-z^2t}{2}}e^{\frac{C_1}{2}},
\end{align}
where in the third and fourth equation, we have integrated with respect to $t$ on both sides and then collected the constants on one side, called $C_1$. The initial value of the ODE is just the initial value of the SDE, $m(0) := \mathbb{E}\left[X_0\right]=X_0 = e^{\frac{C_1}{2}}$. In conclusion, you will end up with the solution:
$$m(t) := \mathbb{E}\left[X_t\right] = X_0e^{-\frac{1}{2}z^2t}$$
In this regard, I believe that you're missing the initial value, $X_0$, in the first part of your statement for $X_t$.
The solution can be verified by observing that the SDE of $X_t$ is the well-known Geometric Brownian Motion (GBM) with solution:
\begin{align}
X_t &= X_0 e^{\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t}\\
&= X_0 e^{\left(\frac{1}{2}(iz)^2-\frac{(iz)^2}{2}\right)t + iz W_t}\\
&= X_0 e^{iz W_t}
\end{align}
where in your scenario, $\mu = \frac{1}{2}(iz)^2$ and $\sigma = iz$. Here, we know that the expected value of a GBM is given by:
\begin{align}
\mathbb{E}\left[X_t\right] &= X_0e^{\mu t}\\
&=X_0e^{-\frac{1}{2}z^2 t},
\end{align}
giving us the same result as the solution of the ODE.
