= Solution
Use <International System of Units>, with $c^{-2}=\mu_0\varepsilon_0$. The <Maxwell equations> with <charge density> $\rho$ and <current density> $\mathbf J$ are
$$
\boxed{\nabla\cdot\mathbf E=\rho/\varepsilon_0,\qquad\nabla\cdot\mathbf B=0,\qquad\nabla\times\mathbf E=-\partial_t\mathbf B,\qquad\nabla\times\mathbf B=\mu_0\mathbf J+c^{-2}\partial_t\mathbf E.}
$$
Locally, or in a suitable contractible region, the zero <divergence> of $\mathbf B$ allows a <magnetic vector potential> $\mathbf B=\nabla\times\mathbf A$. <Faraday's law> then makes $\mathbf E+\partial_t\mathbf A$ <curl>-free, so a <electric potential> can be chosen with
$$
\boxed{\mathbf E=-\nabla\Phi-\partial_t\mathbf A,\qquad\mathbf B=\nabla\times\mathbf A.}
$$
These formulas automatically satisfy the two homogeneous <Maxwell equations>. The local topology assumption rules out obstructions to expressing a closed field as a <gradient> or <curl>.
For any sufficiently smooth scalar $\chi$, the <gauge transformation>
$$
\boxed{\mathbf A\mapsto\mathbf A+\nabla\chi,\qquad\Phi\mapsto\Phi-\partial_t\chi}
$$
leaves both physical fields unchanged, because <curl> of a <gradient> is zero and spatial and time <derivatives> commute. In such a region this accounts for the freedom between potentials for the same fields.
The <Lorenz gauge> is $\nabla\cdot\mathbf A+c^{-2}\partial_t\Phi=0$. Substitution into the sourced <Maxwell equations>, using $\nabla\times(\nabla\times\mathbf A)=\nabla(\nabla\cdot\mathbf A)-\nabla^2\mathbf A$, gives the decoupled <wave equations>
$$
\boxed{(\nabla^2-c^{-2}\partial_t^2)\Phi=-\rho/\varepsilon_0,\qquad(\nabla^2-c^{-2}\partial_t^2)\mathbf A=-\mu_0\mathbf J.}
$$
For example, Gauss's law gives $-\nabla^2\Phi-\partial_t\nabla\cdot\mathbf A=\rho/\varepsilon_0$; the gauge condition turns this into the scalar equation. The remaining gauge freedom preserving the condition has $(\nabla^2-c^{-2}\partial_t^2)\chi=0$.
Put $\theta=\omega t-Kz$ for the given wave. Direct differentiation gives $\nabla\cdot\mathbf E=0$ and $\nabla\times\mathbf E=K\mathbf E$. Integrating <Faraday's law> in time yields the wave's <magnetic field>
$$
\boxed{\mathbf B=\frac K\omega\,\widehat{\mathbf z}\times\mathbf E=\frac{KE}{\omega}(-\sin\theta,\cos\theta,0).}
$$
One could superpose a time-independent <divergence>-free and <curl>-free <magnetic field>; the expression displayed chooses no such extra background. It has zero <divergence> and $\partial_t\mathbf B=-K\mathbf E$, so the magnetic Gauss and Faraday equations hold. Moreover
$$
\nabla\times\mathbf B=\frac{K^2}{\omega^2}\partial_t\mathbf E.
$$
Since the time <derivative> is nonzero, the vacuum <Ampere-Maxwell law> holds precisely when
$$
\boxed{K^2=\omega^2/c^2.}
$$
This verifies all four source-free equations, including both <divergence> constraints. The sign of $\omega/K$ selects the direction of propagation.
A wave has <circular polarization> when, at a fixed spatial point, its transverse electric <vector> has constant nonzero magnitude and rotates uniformly through a circle. Here $E_x^2+E_y^2=E^2$ and the angle changes at rate $\omega$, so \b[the wave is circularly polarized]. Its magnetic <vector> is perpendicular to the electric <vector>, has magnitude $|E|/c$ under the <dispersion relation>, and rotates with it. A handedness label additionally requires a convention for the observer's viewing direction.
The <Lorenz-gauge potentials for a circularly polarized plane wave> give a convenient choice. Take $\Phi=0$ and $\mathbf A=\omega^{-2}\partial_t\mathbf E$. Because $\partial_t^2\mathbf E=-\omega^2\mathbf E$,
$$
-\partial_t\mathbf A=\mathbf E,\qquad
\nabla\times\mathbf A=\omega^{-2}\partial_t(K\mathbf E)=\frac K\omega\widehat{\mathbf z}\times\mathbf E=\mathbf B.
$$
Also $\nabla\cdot\mathbf A=\omega^{-2}\partial_t\nabla\cdot\mathbf E=0$. Together with $\partial_t\Phi=0$, this proves \b[these potentials satisfy the <Lorenz gauge>]. They satisfy the source-free potential <wave equations> because every component has the same <dispersion relation> just obtained.
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