I would like to compute $H_i(S^1\times S^n;\mathbb{Z})$ for $i\geq3$. I believe one approach is to use the Mayer-Vietoris sequence.
If we take the admissible cover consisting of $S^1\times S^n\setminus\{N\}\cong S^1\times D^n \cong S^1$ and $S^1\times S^n\setminus\{S\}\cong S^1$, then we have $$\dots\to H_i(U)\oplus H_i(V)\to H_i(S^1\times S^n)\to H_{i-1}(S^1\times S^{n-1})\to H_{i-1}(U)\oplus H_{i-1}(V)\to\dots $$ Since $(S^1\times S^n\setminus\{N\})\cap (S^1\times S^n\setminus\{S\})\cong S^1\times (S^n\setminus\{N\}\cap S^n\setminus\{S\})\cong S^1 \times S^{n-1}$. Then we get a sequence of isomorphisms $$H_i(S^1\times S^n;\mathbb{Z})\cong H_{i-1}(S^1\times S^{n-1};\mathbb{Z})$$ for $i\geq 3$. We have for $n\geq 3$ $$H_n(S^1\times S^n;\mathbb{Z})\cong H_{2}(S^1\times S^{2};\mathbb{Z})\cong \mathbb{Z}$$ and $$H_{n+1}(S^1\times S^n;\mathbb{Z})\cong H_{2}(S^1\times S^{1};\mathbb{Z})\cong \mathbb{Z}.$$ Now looking at the CW-structure of $S^1\times S^n$, we have that for $2\leq i<n $ and $n+1<i$, $$ H_i(S^1 \times S^n)\cong0. $$ Now, given that $n\geq 3$, the cellular structure looks like $$ \dots \to 0 \to \mathbb{Z} \to \mathbb{Z}\to0\dots 0\to \mathbb{Z} \to \mathbb{Z} $$ The inclusion of $S^1$ into $S^1\times \{*\}\subset S^1\times S^n$ induces a chain map from cellular chain complex of $S^1$ to $S^1\times S^n$. The differential $\partial_1$ of cellular chain complex is trivial since $H_1(S^1)\cong\mathbb{Z}$. So similarly $H_1(S^1\times S^n)\cong\mathbb{Z}$. So we have $$ H_i(S^1\times S^n ; \mathbb{Z})=\begin{cases} \mathbb{Z} & i=0,1,n,n+1\\ 0 & \text{otheriwise} \end{cases} $$
Now I cannot verify if I am correct as I did not find any literature source. Also is there more straight forward solution to this problem?
Here is a more direct approach in my opinion. Either you know this fact, or alternatively it is exercise 2.2.36 in Hatcher: there is a natural isomorphism $$ H_i(S^d \times X; \mathbb{Z}) \cong H_i(X;\mathbb{Z}) \oplus H_{i-d}(X;\mathbb{Z}) $$ for any space $X$ and any natural number $d$. Now we take $X = S^n$ and $d=1$. Since we know the homology groups of $S^n$ your result follows. I hope this is helpful!