You can determine that the direction of the $E$ and $B$ fields are perpendicular to the direction of motion, and you can conclude that $E$ and $B$ are perpendicular. But you can't determine the specific polarization (ie in which direction $E$ is pointing) based on the information given.
This follows since the equations are invariant under rotations about the $z$ axis. Let's say you found the $E$ field pointed in the $x$ direction. But now I can decide to rotate my head by 90 degrees, so that the $E$ field now points in the $y$ direction. The equations are the same in my new frame of reference (as you said, after a rotation the equations are the same). So $E$ pointing in the $y$ direction must also be a valid solution. Therefore, these equations do not have a unique solution for the direction of $E$.
Let's work out a solution to the equations.
First, let's start with the vector form of Maxwell's equations in vacuum (setting $c=1$).
\begin{eqnarray}
\frac{\partial \vec{E}}{\partial t} &=& \nabla \times \vec{B} \\
\frac{\partial \vec{B}}{\partial t} &=& -\nabla \times \vec{E} \\
\nabla \cdot \vec{E} &=& 0 \\
\nabla \cdot \vec{B} &=& 0
\end{eqnarray}
Now we assume a plane wave moving in the $z$ direction, so that $\frac{\partial \vec{E}}{\partial x}=\frac{\partial \vec{E}}{\partial y}=\frac{\partial \vec{B}}{\partial x}=\frac{\partial{\vec B}}{\partial y}=0$. The last two of Maxwell's equations then imply $\frac{\partial E_z}{\partial z}=\frac{\partial B_z}{\partial z}=0$ (the fields have to be perpendicular to the propagation direction).
Now let's assume $\vec{E}$ points in the $x$ direction and $\vec{B}$ points in the $y$ direction. (For linearly polarized waves, we don't lose any generality here since we can always rotate our coordinate system). Then the non-trivial components of the first two equations become
\begin{eqnarray}
\frac{\partial E_x}{\partial t} &=& (\nabla \times \vec{B})_x = \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} = -\frac{\partial B_y}{\partial z} \\
\frac{\partial B_y}{\partial t} &=& -(\nabla \times \vec{E})_y = -\left[\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right]= -\frac{\partial E_x}{\partial z} \\
\end{eqnarray}
Then a solution like $E_x = A \cos\left(k(t-z)\right)\hat{e}_x, B_y=A\cos\left(k(t-z)\right)\hat{e}_y$ will solve the equations.
Another interesting example of a solution is circular polarization
\begin{eqnarray}
\vec{E} &=& \cos(k(t-z)) \hat{e}_x + \sin(k(t-z))\hat{e}_y \\
\vec{B} &=& -\sin(k(t-z)) \hat{e}_x + \cos(k(t-z))\hat{e}_y
\end{eqnarray}
It's a good exercise to see how this solves Maxwell's equations.
