Here is my, admittedly, *ad hoc* way of proving they are distinct. It comes from trying to make it concrete that the graded pieces have different sizes.

First, you better assume that $n\geq 2$ as these rings *are* isomorphic if $n=1$.

Assume, by way of contradiction, there is an isomorphism $\varphi:S\rightarrow R$ where $S$ is the second ring (with the determinant relation) and $R$ is the first ring. Let $f_{ij}=\varphi(y_{ij})$. Thus $\det((f_{ij}))=1$.

Let $I_{0}$ be the ideal of $R$ generated by $2$ and $x_{k}^{2}$ for each $k$. Note that as the $y_{ij}$ generate $S$, we must have that $x_{1}$ occurs in one of the $f_{ij}$ (even modulo $I_{0}$) with non-zero support. Let $I_{1}$ be the ideal generated by the same relations as $I_{0}$, except we add the relation $x_{1}=f_{ij}-x_{1}$. Note that $R/I_{1}$ is isomorphic (naturally) to $\mathbb{Z}[x_2,x_3,\ldots, x_{n^{2}-1}]$.

Now, $x_{2}$ occurs with non-zero support in (a different) $f_{ij}'$ (modulo $I_{1}$) as these polynomials still generate $R/I_{1}$. Create a new ideal $I_{2}$ containing $f_{ij}'$, but for which $R/I_{2}$ looks like $\mathbb{Z}[x_3,\ldots, x_{n^{2}-1}]$.

Repeating this process enough times, we can make make the matrix $(f_{ij})$ have both determinant 1 and 0, modulo an ideal $I$, even though $R/I$ is not the zero ring. This gives you the needed contradiction.