Let $z$ be a complex number such that $|z+2| = |z-2|$ and $\arg\left(\dfrac{z+3}{z-i}\right) = \dfrac{\pi}{4}$. Then $|z|^2$ is equal to:

Given $|z+2| = |z-2|$, the point $z$ lies on the perpendicular bisector of the line segment joining $(-2, 0)$ and $(2, 0)$. This means $z$ lies on the imaginary axis.

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

We are given $\arg\left(\dfrac{z+3}{z-i}\right) = \dfrac{\pi}{4}$. Substituting $z = iy$, we get:

$\dfrac{iy+3}{iy-i} = \dfrac{3+iy}{i(y-1)} = \dfrac{-i(3+iy)}{y-1} = \dfrac{y - 3i}{y-1} = \dfrac{y}{y-1} - i\dfrac{3}{y-1}$

For a complex number $X + iY$ to have an argument of $\dfrac{\pi}{4}$, its real and imaginary parts must be equal and strictly positive. Therefore:

$\dfrac{y}{y-1} = \dfrac{-3}{y-1} > 0$

Since $y \neq 1$, equating the numerators gives $y = -3$.

Checking for positivity: $\dfrac{-3}{-3-1} = \dfrac{3}{4} > 0$, which is valid.

Thus, $z = -3i$.

The value of $|z|^2$ is $|-3i|^2 = 9$.

Answer: $9$