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.

[![enter image description here][1]][1]

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?


  [1]: https://i.sstatic.net/FGmrS.png