Why a tesselation of the plane by a convex polygon of 7 or more sides is not possible?

In a tessellation of the entire plane by convex tiles at least three tiles will meet in each corner of each tile. Therefore the average of inner angles of the tile cannot be higher than 120 degrees. The sum of the inner angles of an $n$-gon is $(n-2)180$. If $n>6$, then their average $180(n-2)/n=180(1-\frac2n)>180(1-\frac26)=120$.

There is a way to cover the entire plane with convex heptagons. The one in the center is regular, then a ring of seven of them are almost regular. The next ring out are already pretty long and narrow, and so on.

From the point of view of your question, the key fact is that these are not all congruent. I'm not sure the word "tessellation" would be correct here.

This construction has been known for a very long time, it is in page 77 in Mathematical Snapshots by H. Steinhaus, probably plenty of other books and web pages. STEINHAUS

Evidently it is also discussed on pages 248-249 of GARDNER

I'm afraid I have been unable to find an image on line. The closest I have found in in $\mathbb H^2,$ see the tiling called (7,3) at HYPERBOLIC and imagine what would happen if the whole thing were stretched radially, in polar coordinates $(r,\theta) \rightarrow \left( \tan \frac{\pi r}{2},\; \theta \right).$ I should admit that I do not really know whether this gives a good approximation of the covering of $\mathbb R^2$ beyond the first three or four rings depicted. Probably not.

As a supplement to Will Jagy's answer, permit me to include two figures. The first image below is from Jay Kappraff's 1990 book, Connections: the geometric bridge between art and science. The second image is from a 1983 paper by Danzer, Grünbaum, Shephard, "Does Every Type of Polyhedron Tile Three-Space?" in Structural Topology 8 (perhaps the source for Kappraff?).