If the electric field strength is $E_x=x, E_y=E_z=0$, then by $\nabla\cdot E=\frac1{\epsilon_0}\rho_e$ where $\rho_e$ is the density of charge, we get $\rho_e=-1$ for any point in the space.
But if $\rho_e=-1$ for any point in the space, then the distribution of charge in space is completely symmetric, so we shouldn't get a $\vec{E}$ which only have $E_x$ component.
I am really confused. Can you explain it for me?
Think about the case where $\rho = 1$ between $x = -x_0$ and $x = x_0$, and two charged planes of $\sigma = -x_0$ at $x = \pm x_0$ . You get $E_x = x$ with this setup inside the interval, and zero field outside. Now let $x_0 \to \infty$, and you get the desired field strength, with some strange charge setup at infinity.
It cannot exist, as it requires an infinite amount of charge and an infinite energy to concentrate this. It also conflicts with average charge neutrality.
Perhaps it is instructive to consider a uniformly charged spherical shell. This is an two dimensional implementation of your idea. The component of the electric field parallel to the surface is indeed zero.
This makes a perfect sense if you consider a plane volume of a finite thickness and the variables inside it.
