If $M_n(R)$ and $M_m(R)$ satisfy the same polynomial identities is it true that $m=n$? Let $K$ be an inifinite field of characteristic different from 2.
The well-known Amitsur-Levitzki theorem states that the algebra $M_n(K)$ satisfy the standard polynomial identity of degree $2n$, 
$$s_{2n}(x_1,\dots,x_{2n})=\sum_{\sigma\in S_{2n}}(-1)^{\sigma}x_{\sigma(1)}\cdots x_{\sigma(2n)}$$
Moreover, it does not satisfy any other identity of degree less than $2n$.
In particular, if $m < n$, $s_{2m}$ is an identity for $M_m(K)$ and is not an identity for $M_n(K)$.
My question is the following: 

If $R$ is a unitary associative noncommutative $K$-algebra that satisfy a polynomial identity, is it true that if $m < n$ then there is an identity of $M_m(R)$ which is not an identity for $M_n(R)$?

In the language of T-ideals, is the inclusion $T(M_n(R))\subset T(M_m(R))$ a proper one?
Equivalently, the PI-equivalence of $M_n(R)$ and $M_m(R)$ imply that $m = n$?
Of course, if the condition that $R$ is a unitary algebra is removed, nilpotent algebras can give counter-examples.
 A: This is not true in general. For example, if $R$ is a free associative algebra of rank $>1$, then $M_n(R)$ does not have nontrivial identities for any $n$. 
Moreover, I believe there are examples of algebras $R$ such that much stronger condition holds: $M_n(R)$ is isomorphic to $M_m(R)$ for some $m \ne n$. This reminds cancellation problems in the commutative setting, though I cannot provide examples of such algebras.
This is true if, for example, if $R$ is finite-dimensional and prime (even not necessary associative). Then we can note that $M_n(R)$ is prime too, pass to the algebraic closure of the base field, and invoke theorem of Razmyslov that finite-dimensional prime algebras over algebraically closed fields are determined by their identities (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30).
EDIT:
I was not careful enough when reading the question, sorry. The question explicitly asks for situation when $R$ is a PI algebra, so the example with a free algebra obviously does not qualify. 
I still think this is not true in the whole generality: probably just the condition of being unitary is to weak, one should demand something like (semi)primeness. 
I can think of two approaches. First, there is a lot of works about identities of tensor product of algebras, and of $M_n(R)$ in particular (typical results: if $R$ satisfies the standard identity of degree $k$, then $M_n(R)$ satisfies the standard identity of some given degree in terms of $n$ and $k$, see e.g. M. Domokos, Eulerian polynomial identities and algebras satisfying a standard identity, J. Algebra 169 (1994), N3, 913-928 DOI:   10.1006/jabr.1994.1317). Second, perhaps one can do something along the lines of Sections 4 and 5 of arXiv:0911.5414.
