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

We have:
$$\frac{z - i}{z + i} = \frac{(x + iy) - i}{(x + iy) + i} = \frac{x + i(y - 1)}{x + i(y + 1)}$$

Multiply numerator and denominator by the conjugate of the denominator:
$$\frac{x + i(y - 1)}{x + i(y + 1)} \cdot \frac{x - i(y + 1)}{x - i(y + 1)} = \frac{(x + i(y - 1))(x - i(y + 1))}{(x + i(y + 1))(x - i(y + 1))}$$

Compute the denominator:
$$(x + i(y + 1))(x - i(y + 1)) = x^2 - i x (y + 1) + i x (y + 1) - i^2 (y + 1)^2 = x^2 + (y + 1)^2$$

Compute the numerator:
$$\begin{align*} (x + i(y - 1))(x - i(y + 1)) &= x(x) + x(-i(y + 1)) + i(y - 1)x + i(y - 1)(-i(y + 1)) \\ &= x^2 - i x (y + 1) + i x (y - 1) - i^2 (y - 1)(y + 1) \\ &= x^2 - i x (y + 1) + i x (y - 1) + (y^2 - 1) \quad (\text{since } i^2 = -1) \\ &= x^2 + (y^2 - 1) - i x (y + 1) + i x (y - 1) \\ &= x^2 + y^2 - 1 - i x (y + 1 - y + 1) \\ &= x^2 + y^2 - 1 - i x (2) \end{align*}$$

So,
$$\frac{z - i}{z + i} = \frac{x^2 + y^2 - 1 - 2ix}{x^2 + (y + 1)^2}$$

The real part is:
$$\text{Re}\left(\frac{z - i}{z + i}\right) = \frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2}$$

We are given that this is $0$, so:
$$\frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2} = 0 \implies x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1$$

But $|z| = \sqrt{x^2 + y^2} = 1$.

Therefore, $|z| = 1$.