This answer proves that four lighthouses is possible. It turns out that it is also possible with five lighthouses.
The first step is to choose a lighthouse, call it $A$, and angle it in such a way that
- there are two lighthouses on either side of the midline of its light
- if you rotated it $180^\circ$, it wouldn't shine on any lighthouse
Below is an illustration:
Why is this possible? Imagine stretching a large rubber around all of the lighthouses, forming a polygon. On the vertices must have an angle of at most $108^\circ$; choose this lighthouse to be $A$. Then rotate $A$ until its midline cuts the other four lighthouses in half. If doing this doesn't fulfill the second bullet point, then turning $A$ around $180^\circ$ will.
Next, we use the two lighthouses below the line to illuminate the rest of the upper half. Let's label a compass with degree marks going counterclockwise so $0^\circ$ is east, meaning $90^\circ$ is north, $180^\circ$ is west, etc.
Of the bottom two lighthouses, let $B$ be the one which is further along in the $342^\circ$ direction, and $C$ be the other. Shine $B$ so that the clockwise edge of its beam points towards $72^\circ$. Shine $C$ so its clockwise edge points east. This will illuminate the upper plane, as shown below:
Note that $A$ and $B$'s beams must overlap because of the second bullet point in how $A$ was chosen, while $B$ and $C$'s beams overlap since $B$ is further along in the $342^\circ$ direction.
Doing the same procedure with the upper two lighthouses illuminates the plane.
As a fun fact, it turns out that if you have any number of lighthouses, and their beam angles add up to $360^\circ$, then they can illuminate the plane. However, the proof is quite mathy.