My idea at the moment goes as follows:
Given an arbitrary projective plane, take the (one of the) longest line(s), and call the number of nodes in this line $n+1$. Note that $n \ge 2$. Separate one node out as the "leader" node. We know that there are at least two other nodes because of rule 3. There are two possibilities for how these nodes add on, given in the picture below (for the $n=4$ case). The leftmost node in each configuration is the "leader."
Now, in the left configuration, there must be a line from 1a to 1b, as well as from 2a to 1b. These cannot both contain 1c, but they must cross the line going from the leader to 1c. Therefore, there must be a "2c" node. This reduces the left configuration to a stronger version of the right configuration.
Now, for the right configuration. There is one line going down from each $1a \dots na$ through each $1b \dots kb$. Each of the lines $2a\to2b \dots na\to2b$ must intersect the line $1a\to1b$, but since they have all already touched, they must each meet $1a\to1b$ at a new node, so we get $n-1$ new nodes. These must all be in different rows, so we now have a total of $n+1$ rows. Sort these rows by length, so the $n+1$th is the shortest. If the $n+1$th row is shorter than the 2nd, then some line that I have already created does not reach the line going from the leader through the $n+1$th row. Therefore, all of the rows $b$ and on must be the same length.
Since n is at least 2, there are at least three rows, so we can look at the last two rows to see that there must be $m^2$ lines between them, where m is the length of the last row (excluding the leader). These must be the same lines as between the first two rows, and it is clear that there are $nm$ lines connecting the first two rows, so we see that $n=m$. Therefore, it is clear that, if this is a finite projective plane, it is an $(n+1)n$ grid, along with a leader node.