I want to show that $S^2$ and $S^3 \times \mathbb CP^\infty$ have isomorphic homotopy groups in each degree. My first approach was to calculate the homotopy group of $\mathbb CP^\infty$ and use the fact that the homotopy group of a product is the product of homotopy groups, but $S^2$ and $S^3$ have such complicated homotopy groups in general that this is not a clever approach.
In calculating the homotopy group of $\mathbb CP^\infty$ we make a fibration $S^1\rightarrow S^\infty \rightarrow \mathbb CP^\infty$. Is there a clever way to make a fibration using $S^2$ and $S^3$ and then get the result using the long exact sequence of the fibration?
As Tyrone points out there is the well-known Hopf fibration $S^1 \to S^3 \to S^2$, which is actually just the restriction of the fibration $S^1 \to S^{\infty} \to \mathbb{CP}^\infty$ to the $2$-skeleton $\mathbb{CP}^\infty_{(2)} = \mathbb{CP}^1 \cong S^2$. From this fibration you find that $\pi_i S^2 \cong \pi_i S^3$ for $i\geq 3$. This result will make your original approach work.
In general, for every $n\in \mathbb{N}\cup\{\infty\}$ there is a tautological fibration $S^1 \to S^{2n+1} \to \mathbb{CP}^n$, and the Hopf fibration is just $n=1$. In particular if $n\geq 1$ then $\pi_2(\mathbb{CP}^n)\cong \pi_1(S^1) \cong \mathbb{Z}$ and $\pi_i(\mathbb{CP}^n) \cong \pi_i(S^{2n+1})$ for $i \geq 3$.