Suppose , $k>1$ is a positive integer.
Consider the equation $$T(n)=T(m)\cdot k^2$$ where $T(n)$ denotes the $n$-th triangle number, so $T(n)=\frac{n(n+1)}{2}$. If we multiply with $8$ and add $1$ , we get $$4n^2+4n+1=(4m^2+4m)k^2+1=(4m^2+4m+1)k^2+1-k^2$$
Hence $$(2n+1)^2-(2m+1)^2\cdot k^2=1-k^2$$
With $u:=2n+1$ and $v:=2m+1$ , this can be written as $$(u-kv)(u+kv)=1-k^2$$
This shows that for every positive integer $k>1$, there are only finite many solutions in positive integers $m$ and $n$. For some $k$, there are $2$ solutions, for example $k=96030$ allows $m=2$ and $m=24007$ , $k=526890$ allows $m=3$ and $m=131722$ , $k=1866294$ allows $m=4$ and $m=466573$.
Is there a $k$, such there are more than $2$ solutions in positive integers ? If yes, what is the smallest such $k$ ?