We are talking about different things, $(x\pm yi)^{|m|}$ is a homogeneous harmonic polynomial of two variables.

Let me give you example. Consider harmonic polynomial of order $3$, containing $z^3$ (there is only one of them, the rest differ by polynomials, which do not contain $z^3$). First of all find it: We need only $z^1$ (as we separate odd and even with respect to $z$; $zxy$ is harmonic by its own, so we take $z^3- az(x^2+y^2)$ for symmetry. Obviously, this is harmonic if $a=\frac{3}{4}$. So, consider

$$z^3-\frac{3}{4}z(x^2+y^2).$$

Plugging $z=\rho\cos(\phi)$, $x=\rho\sin(\phi)\cos(\theta)$, $y=\rho\sin(\phi)\sin(\theta)$ we get

$$

\rho^3\Bigl(\cos^3(\phi) -\frac{3}{4}\cos(\phi)\sin^2(\phi)\Bigr)=\rho^3\Bigl(\frac{7}{4}\cos^3(\phi)-\frac{3}{4}\Bigr).

$$

Here $m=0$, obviously.

Consider harmonic polynomial of order $3$, containing $z^2$. Well it must contain $z^2x $ or $z^2y$, or, better $z^2(x\pm iy)$. This is not harmonic polynomial, to make it correct to $z^2(x\pm yi) + a(x^2+y^2)(x\pm yi)$. Obviously, this is harmonic if $a=\frac{1}{2}$.

$$

\bigl(z^2-\frac{1}{2}(x^2+y^2)\bigr)(x\pm yi) = \rho^3 \bigl(\cos^2(\phi)-\frac{1}[2}\sin^2(\phi)\bigl)\sin(\phi) e^{\pm i\theta}.

$$

Here $m=\pm 1$.

To get to $m=\pm 2$ consider this way $z(x\pm yi)^2$. $m=\pm 3$ consider this way $(x\pm yi)^3$. All four are harmonic without corrections.