A rational triangle is one in which all side lengths are rational numbers.
Question: Can we tile the Euclidean plane with rational triangles that are pairwise non-congruent? No further requirements on the triangles. Their perimeters are not upper or lower bounded.
The answer seems "no" but I have no proof. It is known that there is no tiling of the plane by pairwise non-congruent triangles all of same area and same perimeter (https://arxiv.org/abs/1711.04504) – not sure if that is relevant here.
Further, we can ask if the plane can be tiled by quadrangles and so forth whose sides are all rational (and if the answer is "yes", add further constraints on diagonals etc.).
Best Answer
First answer: The plane can be tiled as requested. First, we tile the plane with equilateral triangles with side lengths $1$. Now each such triangle can be tiled into two rational-sided triangles in infinitely many different ways: Choose a point on a side which dissects it into lengths $x$ and $1-x$. (Sorry, too lazy to produce a picture :-).) Then we get a tiling of the unit triangle into the two triangles of side lengths $1, x, y$ and $1, 1-x, y$, respectively, where $y^2=1-x+x^2$. Setting $x=(1-t^2)/(1+2t)$ for a rational $t$ yields $y=(1+t+t^2)/(1+2t)$. So any rational $0<t<1$ yields a tiling into two rational-sided triangles.
Second answer: There is even a solution with all sides integers! Extend the equilateral triangle with side lengths $m^2-1$ to the one with side lengths $(m+1)^2-1$. So the new piece is a trapezoid with side lengths $m^2-1$, $2m+1$, $(m+1)^2-1$, $2m+1$. Pythagoras tells us that the diagonal has integral length $m^2+m+1$. So we have an obvious inductive procedure to tile the plane. The following picture should illustrate the procedure, starting with the triangle with side lengths $3=2^2-1$:
Third answer: A minor variant to Rosie F's answer, where we arrive one step earlier at a triangle which is similar to the starting one: