Subdivision of triangles into congruent triangles

See http://www.michaelbeeson.com/research/papers/SevenTriangles.pdf and http://www.michaelbeeson.com/research/papers/TriangleTiling1.pdf http://www.michaelbeeson.com/research/papers/TriangleTiling2.pdf http://www.michaelbeeson.com/research/papers/TriangleTiling3.pdf

Beeson conjectures that the cases you list, together with $2n^2$ (a special case of $m^2+n^2$ and $6n^2$ (barycentrically subdivide an equilateral triangle and then split each of the resulting 30-60-90 right triangles into $n^2$ similar triangles) are the only ones possible, without the constraint that the original triangle and the smaller ones be similar. He proves that no subdivision is possible for $n=7$, $11$, $19$, and $23$.

Regarding the first question, yes, those are the only options. Proof can be found in:

S. L. Snover, C. Waiveris, and J. K. Williams. Rep-tiling for triangles. Discrete Math., 91(2):193–200, 1991.