= Solution
In vacuum, the source-free <Maxwell equations> are $\nabla\cdot E=\nabla\cdot B=0$, $\nabla\times E=-\partial_tB$ and $\nabla\times B=c^{-2}\partial_tE$. Put $\vartheta=\omega(t-z/c)$ and denote the field amplitude by $E_0$. The corresponding travelling-wave <magnetic field> is
$$
\boxed{B=\frac{E_0}{c}(-\sin\vartheta,\cos\vartheta,0)}.
$$
Both divergences vanish because the fields have no longitudinal component and no dependence on the transverse coordinates. Direct differentiation gives $\nabla\times E=(\omega E_0/c)(\cos\vartheta,\sin\vartheta,0)=-\partial_tB$ and $\nabla\times B=(\omega E_0/c^2)(-\sin\vartheta,\cos\vartheta,0)=c^{-2}\partial_tE$.
Introducing potentials by $B=\nabla\times A$ and $E=-\nabla\phi-\partial_tA$ automatically satisfies the magnetic-divergence and Faraday equations. The remaining <Maxwell equations> become equations for these potentials, with the usual gauge freedom. For $\omega\ne0$ take $\phi=0$ and
$$
A=\frac c\omega B=\frac{E_0}{\omega}(-\sin\vartheta,\cos\vartheta,0).
$$
Differentiation verifies both potential relations.
The electromagnetic force is $Q(E+v\times B)$. Confinement also supplies a normal reaction force. For motion in a plane of constant $z$, both $v$ and $B$ are tangent to the plane, so their cross product is normal and is balanced by that reaction. It does not enter the in-plane equation $m\ddot r=QE_0(\cos\vartheta,\sin\vartheta)$.
At the fixed plane, the solution with no uniform drift is
$$
r(t)=r_0-\frac{QE_0}{m\omega^2}(\cos\vartheta,\sin\vartheta).
$$
It is circular, with
$$
\boxed{R=\frac{|QE_0|}{m\omega^2}}.
$$
Other initial data can add a constant drift velocity; the question only asks for the possibility of circular motion.
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