I have two matrices $A$ and $B$ that do not commute, i.e., $A B \neq B A$. Thus, if I apply the binomial formula, I get
$$ (A+B)^{n} \neq (B+A)^{n}$$
This is my own deduction. I haven't found any theorem about this property. Do you know if there exists one?
As others have remarked, $(A+B)^n=(B+A)^n$ for arbitrary square matrices $A$, $B$. The problem is with the binomial formula.
If you want to compute, e.g., $(A+B)^3$ for two matrices $A$, $B$ then the distributive law generates $8$ terms as follows: $$(A+B)^3=A^3+A^2B+ABA+AB^2+BA^2+BAB+B^2A+B^3\ .\tag{1}$$ When the matrices commute then you can write each of these terms in the form $A^jB^k$ and then collect similar terms. This leads to the binomial formula $$(A+B)^3=\sum_{k=0}^3{3\choose k}A^{3-k}B^k\ .\tag{2}$$ When the matrices $A$ and $B$ do not commute then $(1)$ is still true, but it is forbidden to convert the $2^n$ terms of $(1)$ into the $n+1$ simpler terms appearing in $(2)$.