Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Given $|z| = \sqrt{2}$, so
$$|z|^2 = x^2 + y^2 = 2.$$

Also, $|z^2 - i| = 1$.
Compute $z^2$:
$$z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy.$$
So,
$$z^2 - i = (x^2 - y^2) + 2ixy - i = (x^2 - y^2) + i(2xy - 1).$$
Therefore,
$$|z^2 - i|^2 = (x^2 - y^2)^2 + (2xy - 1)^2 = 1.$$

Now, we have the system:
$$\begin{align*} x^2 + y^2 &= 2 \\ (x^2 - y^2)^2 + (2xy - 1)^2 &= 1 \end{align*}$$

Let us use polar form: $z = r e^{i\theta}$, $r = \sqrt{2}$, so $z = \sqrt{2} e^{i\theta}$.
Then $z^2 = 2 e^{i2\theta}$, so
$$z^2 - i = 2 e^{i2\theta} - i.$$

Write $2 e^{i2\theta} = 2(\cos 2\theta + i \sin 2\theta)$, so
$$z^2 - i = 2\cos 2\theta + i(2\sin 2\theta - 1).$$

Therefore,
$$|z^2 - i|^2 = (2\cos 2\theta)^2 + (2\sin 2\theta - 1)^2 = 1.$$
Expand:
$$\begin{align*} 4\cos^2 2\theta + (4\sin^2 2\theta - 4\sin 2\theta + 1) &= 1 \\ 4\cos^2 2\theta + 4\sin^2 2\theta - 4\sin 2\theta + 1 &= 1 \\ 4(\cos^2 2\theta + \sin^2 2\theta) - 4\sin 2\theta + 1 &= 1 \\ 4(1) - 4\sin 2\theta + 1 &= 1 \\ 4 - 4\sin 2\theta + 1 = 1 \\ 4 - 4\sin 2\theta = 0 \\ 1 - \sin 2\theta = 0 \\ \sin 2\theta = 1 \end{align*}$$

So $2\theta = \frac{\pi}{2} + 2\pi k$, $k \in \mathbb{Z}$.
Thus $\theta = \frac{\pi}{4} + \pi k$.

Therefore, the solutions are
$$z = \sqrt{2} e^{i\left(\frac{\pi}{4} + \pi k\right)}$$
for $k \in \mathbb{Z}$.

Explicitly, for $k = 0$:
$$z_1 = \sqrt{2} e^{i\frac{\pi}{4}} = \sqrt{2}\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) = \sqrt{2}\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = 1 + i.$$
For $k = 1$:
$$z_2 = \sqrt{2} e^{i\left(\frac{5\pi}{4}\right)} = \sqrt{2}\left(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}\right) = \sqrt{2}\left(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = -1 - i.$$

Thus, the solutions are $z = 1 + i$ and $z = -1 - i$.