This is a pretty simple result in mathematics, namely that there are exactly three tessellations of the plane by regular polygons. If I'm not mistaken, I can show this is true using some fairly simple geometry and integer factorization.
Any point in a plane tessellated must have equal vertex angles at the point. We can divide 360 in such a way that we choose the possible angles to be those less than 180 (because this is the only case which really makes sense, lest we divide it simply into a half-plane), and note the following:
First, any non-trivial n-gon must have an interior angle of at least 60 degrees, that is a triangle. So we note that the only values we care about amongst the following list of possible angle values are those greater than or equal to 60 and less than 180. By factorizing 360, and using simple combinatorics you have this:
1,2,4,5,8,9,10,12,15,18,20,24,30,36,40,60,90,120
as the possible interior angles of your tessellating polygons, note by the previous we can reduce this list now to 60, 90 and 120.
And we now refer to the fact that a regular n-gon has interior angle equal to (n-2)*180 *(1/n) and this must give:
(n1 - 2)*3 = n1
(n2 - 2)*2 = n2
(n3 - 2)*(3/2) = n3
For which we get the solutions n1 = 3, n2 = 4, n3 = 6.
This establishes that the only possible regular tessellations are by equilateral triangles, squares and regular hexagons.
EDIT: I apologize if I forgot some possible angle values when writing them down, I was trying to compute them fairly quickly, and might have missed some particular combination of factors, either way it wouldn't affect the outcome.