Tiling with incommensurate triangles Say that two triangles are incommensurate if they do not
share an edge length or a vertex angle, and their areas differ.
Suppose you'd like to tile the plane with pairwise incommensurate triangles.
I can think of at least one strategy.
Spiral out from an initial triangle
in the pattern depicted below, with the red extensions chosen
to avoid length/angle/area coincidences with all previously
constructed triangles.

          


          

The central $\triangle$ is slightly non-equilateral.
All $\triangle$s incommensurate.


It seems clear that this approach could work, although it might not
be straightforward to formalize to guarantee incommensurate triangles.
Which brings me to my question:


Q.
  What is a scheme that details a lattice tiling—all vertices at
  points of $\mathbb{Z}^2$—composed of pairwise incommensurate triangles?

This requires a more explicit design that effectively describes the triangle
corner coordinates in a way that makes it evident that no
lengths/angles/areas are duplicated.
Without such a clear description, it is not even immediately evident (to me) that
it is possible.
The same question may be asked for incommensurate simplex tilings
with vertices in $\mathbb{Z}^d$.

See also: Tiling the plane with incongruent isosceles triangles.
 A: Edit: This is a solution for non-commensurate side lengths only. The question also asks for non-commensurate angles, which this example does not provide.
I follow your approach. I start with a triangle with points in $\mathbb{Z}^2$ and add three incommensurate triangles, which leads to a new large triangle that is congruent to the first one and has points in $\mathbb{Z}^2$. We can repeat this process to tile the plane.
The first points are $A(0,0), B(2,0), C(2,1)$ as in the following image.

We now add three new points $A'(-4,0), B'(2,-14), C'(24,12)$ on the rays. We note:
(1) All new side lengths are strictly larger than the side lengths of the triangle $A,B,C$.
(2) All new side lengths are distinct (exercise).
(3) The new triangle $\Delta C'A'B'$ is congruent to $\Delta ABC$, in fact there is a right angle at $A'$ and the side $A'C'$ is double as large than $A'B'$.
The new arrangement of the four triangles is incommensurate by (1) and (2). 
We can now repeat this process. Since (3) holds, we can use the congruency from $\Delta ABC$ to $\Delta A'B'C'$ to get the next larger triangle. By (1), all new lengths are larger than all old lengths and the new lengths are also distinct by (2). Moreover all points lie on $\mathbb{Z}^2$.
We can fill up all of $\mathbb{R}^2$ with incommensurate triangles.
Interesting follow up questions:


*

*Is there a tiling in which the size of the triangles is bounded?

*How fast do the lengths grow?

