For any $x,y\geq 0$, there is a regular submanifold of $\mathbb{R}^{x+y+1}$ which is a diffeomorphism to $S^x\times S^y.$
My idea is what if we let $S$ to be a regular sub manifold of $\mathbb{R}^2$ and start from there. By the definition of regular submanifold, there exists a chart $(U,\phi)$ of $\mathbb{R}^2$ such that $S\cap \mathbb{R}^2\{x\in\mathbb{R}^2\vert x_{\dim(S)+1}=\cdots=m_{\dim(\mathbb{R}^2)}=0\}$. But unsure how to proceed. To show there is a regular sub manifold which is a diffeomorphism to $S^m\times S^n$, maybe it is not a good place to start with $S$. Instead we should start with arbitrary regular sub manifold of $\mathbb{R}^{x+y+1}$ and somehow show a diffeomorphism relation. Any help is greatly appreciated.
You can embedd $\mathbb{S}^{x}\times\mathbb{R}$ into $\mathbb{R}^{x+1}$ as follows. First notice that $f:\mathbb{S}^{x}\times\mathbb{R}^+\rightarrow \mathbb{R}^{x+1}$ given by $f(x,t)=tx$ is an embedding. Then, since $\mathbb{R}^+\cong\mathbb{R}$ diffeomorphic via the natural logarithm, you are done.
Next, as composition of embeddings are still embeddings, and $\mathbb{S}^y$ naturally sits inside $\mathbb{R}^{y+1}$, you get a chain of embeddings $$\mathbb{S}^{x}\times\mathbb{S}^y\hookrightarrow\mathbb{S}^{x}\times\mathbb{R}^{y+1}\cong(\mathbb{S}^{x}\times\mathbb{R})\times\mathbb{R}^y\hookrightarrow\mathbb{R}^{x+y+1}.$$ The same arguments proves that there is an embedding $\mathbb{S}^{n_1}\times\ldots\times\mathbb{S}^{n_k}\hookrightarrow\mathbb{R}^{n_1+\ldots+n_k+1}$.