Let $k\geq 1$ be given. Consider the following statement:
For all (non equilateral) triangles (represented by 3 points in $\mathbb R^2$) and for all $k$-colorings of $\mathbb R^2$ there exists a monochromatic congruent image of this triangle.
I think for $k=2$ this is believed to be true, but what about $k\geq 3$? Is it wrong if I allow more than $2$ colors?
It is not true for $k \geq 3$. Color the plane with horizontal strips of unit height and alternating three colors (say, red blue green). Now look at an isosceles right triangle with leg length $1.99$.
Suppose there is a congruent monochromatic triangle, say red. Since all other red strips are at a distance at least $2$ away from the strip containing the right angled vertex, and the other vertices are only $1.99$ away from the right angled vertex, all the vertices of the triangle must lie in the same strip. Hence the entire triangle lies in the same strip, and so does its incircle. But the diameter of the incircle of the triangle is more than $1.165$, and specifically more than $1$, the height of the strip, a contradiction.