For any ring $R$ and any $n\geq 1$, $R$ is Morita equivalent to $\text{Mat}_n\ R$ through the $R-\text{Mat}_n\ R$-bimodule $R^{\oplus n}$. In particular, the relation $\sim$ on central simple algebras induced by (i) $A\sim B$ for $A\cong B$, and (ii) $A\sim\text{Mat}_n\ A$ for $n\geq 1$, is coarser than the relation $\sim_{\text{Mor}}$ of Morita equivalence.
Conversely, suppose $A$ and $B=\text{Mat}_m\ D$ are CSA, with $D$ a division algebra, and assume $A$ and $B$ are Morita equivalent. Then, since $B\sim_{\text{Mor}} D$, also $A$ and $D$ are Morita equivalent. An equivalence ${\mathscr F}: A\text{-Mod}\cong D\text{-Mod}$ sends the (Noetherian) regular left $A$-module $_A A$ to a $D$-module of the form $_DD^{\oplus n}$ for some $n$ (all $D$-modules have a basis), hence $$A^{\text{op}}\cong\text{Hom}_{A\text{-Mod}}(_AA,\ _AA)\xrightarrow{\mathscr F}\text{Hom}_{D\text{-Mod}}(_DD^{\oplus n},\ _DD^{\oplus n})\\\quad\cong\text{Mat}_n(\text{Hom}_{D\text{-Mod}}(_DD,_DD))\cong\text{Mat}_n(D^{\text{op}})\cong\text{Mat}_n(D)^{\text{op}},$$
the last isomorphism by virtue of the transpose map. Hence $A\cong\text{Mat}_n\ D$, so $A\sim \text{Mat}_n\ D\sim D\sim B$.