Theorem: For $n\ge 2$, if an $h\times w$ rectangle can be tiled by right-handed copies of $T_n$, then either $h$ or $w$ is a multiple of $n$, and either $h$ or $w$ is a multiple of $n+1$.
Provided $\min(h,w)\ge n^2-n$, the converse is also true.
Therefore, if a rectangle can be tiled by $T_n$, then its area is a multiple of $n(n+1)$. Furthermore, letting $R_n$ be the $n\times (n+1)$ rectangle, both $T_n$ and and $R_n$ can tile the same sufficiently large rectangles.
In what follows, I will assume $n\ge 3$. My answers to the previous question gave a sketch of the proof for $n=2$, which turns out to be much harder.
We will use the tiling group invented by Conway and Lagarius [1]. The two tiles are described by the boundary words $x^ny^n(x^{-1}y^{-1})^n$ and $y^nx^n(y^{-1}x^{-1})^n$, so the tiling group is
$$
G:= \langle x,y \mid x^ny^n(x^{-1}y^{-1})^n=y^nx^n(y^{-1}x^{-1})^n=1\rangle
$$
The boundary word of a $h\times w$ rectangle is $x^wy^hw^{-w}y^{-h}$. If the rectangle can be tiled, then its boundary is the identity in $G$, so we will show that $x^wy^hx^{-w}y^{-h}=1$ implies the divisibility conditions.
It suffices to consider the image of $G$ in a certain permutation group. Both $x$ and $y$ will be realized as permutations of $(\Bbb Z/n\Bbb Z)^2$. For $x$, we use the permutation $(i,j)\mapsto (i+1,j)$, while the image of $y$ is the permutation defined by
$$
\begin{align}
y(0,j) &= (0,j)
\\
y(1,j) &= (1,j+2),
\\
\text{For } i\ge 2,\;\; y(i,j)&=(i,j+1)
\end{align}
$$
It should be clear that $x^n=y^n=1$. With some thought, you can show $(x^{-1}y^{-1})^n=(y^{-1}x^{-1})^n=1$ as well. This further implies that both of the relations defining $G$ are satisfied, so this defines a valid homomorphism. However, $x^w$ does not commute with $y^h$, except in the trivial cases where $n\mid h$ or $n\mid w$. To see this, it suffices to consider the images of $(0,0)$ and $(-1,0)$.
$$
\begin{align}
w\neq 0,1\implies (y^h\circ x^w)(0,0)&=(w,h),&\qquad (x^w\circ y^h)(0,0)&=(w,0)
\\
w=1\implies (y^h\circ x^w)(-1,0)&=(0,0),&\qquad (x^w\circ y^h)(-1,0)&=(0,h)
\end{align}
$$
We have shown that being able to tile the $h\times w$ rectangle implies that $n\mid h$ or $n\mid w$, so we are half way done. For the other half, it helps to use the following alternate presentation of $G$, which attained by conjugating the two relations in $G$:
$$
G = \langle x,y \mid x^{n+1}y^{n+1}(y^{-1}x^{-1})^{n+1}=y^{n+1}x^{n+1}(x^{-1}y^{-1})^{n+1}=1\rangle
$$
This shows shows that we can represent $x$ and $y$ as permutations of $(\mathbb Z/(n+1)\mathbb Z)^2$, in a way exactly analogous to before, which proves that $x^w$ and $y^h$ do not commute unless $(n+1)\mid h$ or $(n+1)\mid w$.
A harder related question to yours is this; for which integers $n>m$ can $T_n$ be tiled by right-handed copies of $T_m$? The case $m=2$ was analyzed by Conway and Lagarius; they showed that $T_n$ can be tiled by $T_2$ if and only if
$$
n\equiv 0,2,9,\text{ or }11\pmod{12}
$$
It is a fun puzzle to try to tile $T_{12}$ with these triangles. Later, Thurston [2] showed that $n> m\ge 3$ implies that $T_n$ cannot be tiled by $T_m$.
The permutation representation looks like it came out of nowhere, so let me add some motivation. When presented with a complicated group like $G$, a natural first step is to look for a large normal subgroup, $N$, and see if the quotient $G/N$ is sufficient. This is exactly what Thurston did in $[2]$, for (approximately) this same group. He found that $N=\langle x^{n(n+1)},y^{n(n+1)},(xy)^{n(n+1)}\rangle $ is a central subgroup, and that the quotient splits as a direct product of two von Dyck groups. For integers $\ell,m,n\ge 2$, the von Dyck group is defined to be
$$
D(\ell,m,n)=\langle x,y\mid x^\ell=y^m=(xy)^n=1\rangle,
$$
and Thurston proved
$$
G/N\cong D(n,n,n)\oplus D(n+1,n+1,n+1)
$$
The von Dyck group has a geometric realization; given a triangle whose angles are $\pi/\ell,\pi/m,$ and $\pi/n$, the von Dyck group consists of symmetries of the plane generated by rotations by $2\pi/\ell,2\pi/m$, and $2\pi/n$ (resp.) through the vertices of this triangle. The plane has either spherical, Euclidean, or hyperbolic geometry, according to the whether the sum of the triangle angles is greater than, equal to, or less than $2\pi$. Therefore, we reduce the problem to two questions; how can you prove that $x^w$ and $y^h$ do not commute in $D(n,n,n)=\langle x,y\mid x^n=y^n=(xy)^n\rangle $, and the same question when $n$ is replaced with $n+1$?
When $n=3$, $D(3,3,3)$ consists of Euclidean symmetries, and it is easy to visualize them and prove $x^w$ and $y^h$ do not commute. (The Cayley graph of $D(3,3,3)$ is the tessellation of hexagons and triangles mentioned in the other answer). For larger $n$, the plane is hyperbolic; undoubtedly someone familiar with hyperbolic geometry could easily prove these symmetries did not commute, but I needed some sort of representation. After googling for representations of triangle groups, I came across this paper, which inspired me to look for permutations of a discrete $n\times n$ grid.
Sources:
J.H Conway, J.C Lagarias, Tiling with polyominoes and combinatorial group theory, Journal of Combinatorial Theory, Series A, Volume 53, Issue 2, 1990, Pages 183-208, https://doi.org/10.1016/0097-3165(90)90057-4.
Thurston, William P. Conway’s Tiling Groups. The American Mathematical Monthly, vol. 97, no. 8, 1990, pp. 757–73. JSTOR, https://doi.org/10.2307/2324578.
Best Answer
Indeed, you can show that $$ x^4y^2x^{-4}y^{-2}=1. $$ Since the boundary of a $2\times 4$ rectangle is trivial , the tiling group is insufficient to prove that $4\mid n$ and $4\mid m$ are necessary, unfortunately. This was shown in Tile Homotopy Groups by Michael Reid, Corollary $6.6$, and my proof closely follows his.
Lemma: $xyx^{-1}y^{-1}$ commutes with all words of even length.
Proof: We first show that $xyx^{-1}y^{-1}$ commutes with $yx$. \begin{align} (yx)(xyx^{-1}y^{-1})(yx)^{-1}(xyx^{-1}y^{-1})^{-1} &= yx^2yx^{-1}y^{-2}x^{-1}\\ &= yx^2\cdot \color{blue}1\cdot yx^{-1}y^{-2}x^{-1}\\ &= yx^2\cdot \color{blue}{x(y^{-1}x^{-1})^2y^3xy^{-1}}\cdot yx^{-1}y^{-2}x^{-1}\\ &= yx^3 (y^{-1}x^{-1})^2yx^{-1}\\ &=1 \end{align} The blue substitution follows since $\color{blue}{x(y^{-1}x^{-1})^2y^3xy^{-1}}$ is the boundary of a tile. The same goes for the last step.
Similarly, you can show $xy$ commutes with $xyx^{-1}y^{-1}$. Indeed, the commutator of $(xyx^{-1}y^{-1})^{-1}$ and $(xy)$ is equal to $$ \begin{align} (xyx^{-1}y^{-1})^{-1}(xy)(xyx^{-1}y^{-1})(xy)^{-1}= yx^2yx^{-1}y^{-2}x^{-1}, \end{align} $$ and the latter was the second expression in our previous computation.
There is an automorphism of $G$ determined by $90^\circ$ clockwise rotation of the plane, given by $x\mapsto y^{-1}$ and $y\mapsto x$. Applying this to the fact that $xy$ commutes with $xyx^{-1}y^{-1}$, we get that $y^{-1}x$ commutes with $y^{-1}xyx^{-1}$, which after conjugating by $y$ implies $xy^{-1}$ commutes with $xyx^{-1}y^{-1}$. Since $$ x^2=(xy^{-1})(yx)\qquad y^2=(xy^{-1})^{-1}(xy), $$ we get $x^2$ and $y^2$ commute with $xyx^{-1}y^{-1}$. The fact that $xyx^{-1}y^{-1}$ commutes with all other two letter words easily follows. $\square$
Finally, we prove that $x^4y^2x^{-4}y^{-2}=1$. $$ \begin{align} 1&= x^4y^3x^{-1}y^{-1}x^{-2}y^{-1}x^{-1}y^{-1}\tag{see picture} \\&=x^4y^2x^{-1}(\color{red}{xyx^{-1}y^{-1}})(\color{orange}{x^{-2}y^{-1}x^{-1}})y^{-1} \\&=x^4y^2x^{-1}(\color{orange}{x^{-2}y^{-1}x^{-1}})(\color{red}{xyx^{-1}y^{-1}})y^{-1}\tag{Lemma} \\&=x^4x^2x^{-4}y^{-2}. \end{align} $$ The first equation follows because the RHS is the boundary of a region that can be tiled with T-tetrominos:
It is very cool that you can prove group relations with tiling pictures!
Going further, we can show that $G$ tells us nothing that a coloring argument already told us. Let $G_\text{cyc}$ be the subgroup of $G$ consisting of words that determine a closed path in the plane; this is the only part we care about for tiling arguments. You can prove that $$ G_\text{cyc}\cong \mathbb Z/8\mathbb Z $$ The isomorphism is as follows. Color the grid like a checkerboard, and given a word in $G$, let $B$ be the number of black squares enclosed. To be clear, this is the sum of the winding numbers of the path around the centers of all black squares. Define $W$ similarly for the white squares enclosed. The isomorphism is $g\mapsto (B+5W)\pmod 8$. But we did not need the non-commutative tiling group to tell us this; it follows because each tetromino covers either 3 black squares and 1 white square, or 1 black and 3 white squares.
Reid, Michael. (2003). Tile homotopy groups. Enseignement Mathématique. 49. 123-155. https://doi.org/10.5169/seals-66684
Walkup, D. W. (1965). Covering a Rectangle with T-Tetrominoes. The American Mathematical Monthly, 72(9), 986–988. https://doi.org/10.2307/2313337