Tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$ [duplicate]

I am trying to show that the tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$. This is from an exam, where there is a hint stating that this is more than showing that $TS^2$ is non-trivial.

I know how to show the hairy ball theorem, according to which $TS^n$ is non-trivial iff n is even.

I also know that a vector bundle $\pi:E\rightarrow M$ of rank $m$ on a smooth manifold $M$ is trivial (by definition) iff there exists a diffeomorphism $f:E\rightarrow M\times \mathbb{R}^m$ such that for every $p\in M$, $f$ induces a vector space isomorphism $f:\pi^{-1}(p)\rightarrow \{p\}\times \mathbb{R}^m$.

So I see that showing that $TS^2$ is non-trivial only guarantees that $TS^2$ is not diffeomorphic to $S^2\times \mathbb{R}^2$ via a diffeomorphism satisfying the property above, but it is not enough to conclude that there isn't any diffeomorphism.

How can I show this then?

Here's a proof that relies on a theorem of Kervaire with a somewhat-involved proof (I think the particular case of $n=k=2$ is probably much easier, but I couldn't find such a proof to confirm my suspicions.)

Theorem: The normal bundle to an $n$-sphere embedded inside $\Bbb R^{n+k}$ is trivial if $k> \frac{n+1}2$

This implies any 2-sphere embedded in $S^2 \times \Bbb R^2$ has a tubular neighborhood isomorphic (as fiber bundles) to $ S^2 \times \Bbb R^2$, since $S^2 \times \Bbb R^2$ embeds as an open manifold of $\Bbb R^4$. If $f:TS^2\rightarrow S^2 \times \Bbb R^2$ is a diffeomorphism, then the image of the zero section has a trivial tubular neighborhood $N$ by the above disscussion, but then $f^{-1}(N)$ is a trivial tubular neighborhood of the zero section in $TS^2$.

References:

• Massey's On the Normal Bundle of a Sphere Imbedded in Euclidean Space
• Kervaire's An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant