= Solution
For the usual hollow <rectangular electromagnetic waveguide>, let the propagation axis be $z$ and the interior be $0<x<a$, $0<y<b$. A <transverse magnetic polarization> has $B_z=0$ and $E_z\ne0$. At a <perfect conductor> wall the tangential <electric field> vanishes. Since $E_z$ is tangential to all four side walls, it satisfies a homogeneous <Dirichlet boundary condition> around the rectangle.
Separating the transverse <Helmholtz equation> therefore gives the <transverse magnetic mode of a rectangular waveguide>
$$
E_z=E_0\sin\left(\frac{n\pi x}{a}\right)\sin\left(\frac{m\pi y}{b}\right)e^{i(k_z z-\omega t)},
\qquad n,m=1,2,\ldots.
$$
A zero value of either index makes $E_z$ identically zero, rather than producing a nonzero field uniform along that direction. The transverse electric and <magnetic fields> of this TM family are obtained from transverse <derivatives> of $E_z$, so they also vanish. Thus \b[both TM indices must be nonzero: neither $\mathrm{TM}_{n0}$ nor $\mathrm{TM}_{0m}$ exists] in this rectangular hollow-guide model. In contrast, transverse electric modes obey a <Neumann boundary condition> on $B_z$ and may have one zero index. The claim is about this <waveguide> geometry, not all possible guides.
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