Complex notation of monochromatic planar waves in Griffith's The definition of planar waves in Griffith's Electrodynamics textbook is given as:
The waves are travelling in the z direction and have no x or y dependence; these are called plane waves, because the fields are uniform over every plane perpendicular to the direction of propagation.
Along with this image:

Now generally
$\mathbf E(x,y,z,t) = \mathbf E_0(x,y,z)e^{i(\mathbf {k.r}-\omega t)}$
So that means for plane waves,
$\mathbf E(z,t) = \mathbf E_0(z)e^{i(kz-\omega t)}$
But in the picture that's given, clearly the wave has x dependence. I mean how can a wave be only one dimensional like F(z)? Also, how is the field uniform? (as its changing with time?)
 A: *

*Scalar field

A scalar field $\psi (\mathbf{r}, t)$ is a function that assigns a value to each point of space (and time).
Take for example, $\psi (\mathbf{r},t) = z^3 t$. It doesn't depend on $x$ nor $y$ so, for example, $\psi(x=1,y=2,z=3,t=4) = \psi(x=-\pi,y=0,z=3,t=4) = 108$. For all points in the plane $z=3$, the function will have the same value $\psi(z=3)=27t$ . The scalar field changes in time but in the same way for each plane.
I don't understand what you mean by "1 directional". Not depending on some variables doesn't changes the mathematical object you are dealing with: it's still a function with dimensions $\mathbb{R}^{3+1}\rightarrow\mathbb{R}$.


*Planar Wave

$$\psi(\mathbf{r},t) = \psi_0 e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$
$$\mathbf{E}(\mathbf{r},t) = \mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$
$$\mathbf{E}(\mathbf{r},t) = E_{x} e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t +\phi_{x})} \mathbf{\hat{x}} + E_{y} e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t +\phi_{y})} \mathbf{\hat{y}}$$
are all examples of planner waves.
It doesn't matter if you have a scalar field or a vector field, what makes a wave planar is the fact that all the point in planes perpendicular to the propagation direction have the same value. That value is a scalar or a vector depending on the type of wave.
A: 
But in the picture that's given, clearly the wave has $x$ dependence.

When Griffiths says this, he means that the field value doesn't depend on $x$.  So (for example), the value of the electric field at $x = 0$, $y = 12$, $z = 2$, $t = 0$ is always the same as the electric field at $x = 4$, $y = 12$, $z = 2$, $t = 0$, or the electric field at $x = -1000$, $y = 12$, $z = 2$, $t = 0$.

Also, how is the field uniform? (as its changing with time?)

The statement applies to a particular moment in time.  You're correct that on each of the planes perpendicular to the direction of propagation, the field value will depend on time.  What Griffiths is saying is that at a particular moment in time, the field is has the same value (direction & magnitude) at every point on a given plane.

I mean how can a wave be only one dimensional like F(z)?

If I understand what you're asking here correctly, the answer is that we've chosen to look for a solution to Maxwell's equations that varies only with respect to $z$ (and $t$.)  A priori, we don't know whether such a solution exists;  what Griffiths does in the chapter you're referring to is to show that such solutions do exist and to derive their properties.
