Well, if you tile the plane by congruent regular polygons, there must be $n$ polygons meeting at each vertex. Thus the interior angles of each polygon must be $2\pi/n$, for some positive integer $n$.
For $n=3$, we get polygons with angles of $2\pi/3$, which are regular hexagons. This tiling has three regular hexagons meeting at each vertex.
For $n=4$, we get polygons with angles of $2\pi/4 = \pi/2$, which are squares. This tiling has four squares meeting at each vertex.
For $n=5$, the polygons would need to have angles of $2\pi/5$. This is not possible for a regular polygon.
For $n=6$, the polygons would need to have angles of $2\pi/6 = \pi/3$, which are equilateral triangles. This tiling has six triangles meeting at each vertex.
For $n>6$, the polygons would need to have angles less than $\pi/3$, which is impossible.
Edit: As Blue points out below, this argument neglects tilings such as the brick wall tiling, where vertices of one polygon meet edges of another. See Steven Stadnicki's comment for the resolution of this case.