= Solution
Use the <Minkowski metric> $\eta=\operatorname{diag}(-1,1,1,1)$ and define the <electromagnetic field tensor> by
$$
F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu.
$$
As a rank-two <Lorentz tensor>, it transforms under a <Lorentz transformation> as
$$
\boxed{F'^{\mu\nu}(x')
=\Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma}F^{\rho\sigma}(x).}
$$
Two independent quadratic <electromagnetic field invariants> are
$$
I_1=\frac12F_{\mu\nu}F^{\mu\nu},
\qquad
I_2=\frac12F_{\mu\nu}\widetilde F^{\mu\nu},
\qquad
\widetilde F^{\mu\nu}=\frac12\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.
$$
With $A^\mu=(\phi/c,\mathbf A)$, $\mathbf E=-\nabla\phi-\partial_t\mathbf A$, and $\mathbf B=\nabla\times\mathbf A$, the components are
$$
F^{\mu\nu}=
\begin{pmatrix}
0&E_x/c&E_y/c&E_z/c\\
-E_x/c&0&B_z&-B_y\\
-E_y/c&-B_z&0&B_x\\
-E_z/c&B_y&-B_x&0
\end{pmatrix}.
$$
Direct contraction gives
$$
\boxed{I_1=\mathbf B^2-\frac{\mathbf E^2}{c^2}=S,}
\qquad
\boxed{I_2=-\frac{2}{c}\mathbf E\mathbin\cdot\mathbf B=-\frac{2}{c}T,}
$$
where the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are <Lorentz scalar>[Lorentz scalars].
For a constant uniform field, the <relativistic Lorentz force> is a constant-coefficient <linear ordinary differential equation>. Its solution is the <matrix exponential>
$$
\boxed{u^\mu(\tau)=
\left[\exp\left(\frac{q\tau}{m}F\right)\right]^\mu{}_{\nu}u^\nu(0)
=\sum_{n=0}^{\infty}\frac1{n!}
\left(\frac{q\tau}{m}\right)^n
(F^n)^\mu{}_{\nu}u^\nu(0).}
$$
Now $S=T=0$ implies that the field is a <null electromagnetic field>:
$$
\mathbf E\perp\mathbf B,
\qquad |\mathbf E|=c|\mathbf B|.
$$
The stated axes allow us to choose
$$
\mathbf B=B\mathbf e_z,
\qquad
\mathbf E=cB\mathbf e_x.
$$
Put $\Omega=qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ is
$$
N=\begin{pmatrix}0&1&0\\1&0&1\\0&-1&0\end{pmatrix},
\qquad N^3=0.
$$
The particle starts with <four-velocity> $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and gives
$$
u^0=c\left(1+\frac{\Omega^2\tau^2}{2}\right),
\qquad
u^x=c\Omega\tau,
\qquad
u^y=-\frac{c\Omega^2\tau^2}{2},
\qquad u^z=0.
$$
Integrating with the particle initially at the origin gives the <relativistic trajectory in a constant null crossed field>
$$
x=\frac{c\Omega\tau^2}{2},
\qquad
y=-\frac{c\Omega^2\tau^3}{6}.
$$
Eliminating $\tau$ yields
$$
\boxed{y^2=Ax^3,
\qquad A=\frac{2qB}{9mc}.}
$$
If the original positive $x$ direction is opposite to $\mathbf E$, both the sign of $x$ and the displayed signed coefficient reverse.
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