So, we must choose n such that n-2 | 2n.
This looks already as if there are only few possibities. First note that n must be larger than 2. (Otherwise we don't even have a polygon).
n=3 gives 1|6 which is true, and k = 6/1 = 6
n=4 gives 2|8 which is true, and k = 8/2 = 4
n=5 gives 3|10 which is NOT true
n=6 gives 4|12 which is true, and k=12/4 = 3
How does it go on?
Well, the next quotient will be smaller than 3, so it must be 2, but then this would mean that just two corners n-gon fill an arc of 2Pi and this corner should be Pi, but this does not happen for a finite n.
Therefore, there are no other regular n-gons that tesselate the plane.
You can also look at solving the same equation (*) for n, we get:
n = 2k/(k-2)
So k should satisfy k-2 | 2k
(i) k-2 = 1 or
(ii) k-2 = 2 or
(iii) k-2 is odd and k-2|k
(iv) k-2 is even and (k-2)/2 | k
Ad (i) k=3 and n=6
Ad (ii) k=4 and n=4
Ad (iii) k=2i+1 and 2i-1 | 2i+1. It is clear that if i>1 then (2i+1)/(2i-1) < 2, so this leaves no solutions
Ad (iv) k=2i and (i-1)|2i eq. i=2 OR i=3 only. i=2 gives k=4 and we had this already. i=3 gives k=6 and n=3.