You can prove the Fundamental Theorem of Algebra using Sylow 2-subgroups. The sketch of the proof is as follows:

1. Take a (WLOG) normal extension $k$ of $\mathbb{R}$ of degree $n$ and a 2-subgroup $H$ of $G=$Gal$(k/\mathbb{R}$). Then $[\text{Fix}(H):\mathbb{R}]=\vert G \vert / \vert H \vert $ and hence since $H$ is a 2-group $n$ must be odd and hence trivial. So $G=H$, i.e. $\vert G \vert = 2^m$
2. Therefore, by taking a subgroup $N$ of $G$ of index $2$ show that Fix$(N)/\mathbb{R}$ is $\mathbb{R}(\theta)$ for a negative square root $\theta$ and hence that Fix$(N)=\mathbb{C}$.
3. Then by a quick argument Gal$(k/\text{Fix}(N))=$Gal$(k/\mathbb{C})=1$ and hence $K=\mathbb{C}$. Hence $\mathbb{C}$ is the only extension of $\mathbb{R}$ and the FTA follows immediately.

That said, students will most likely encounter Sylow p-groups before Galois Theory. But that's not really an argument against providing the proof of FTA as a 'spectacular' application - especially if students are familiar with some basic field theory. The Galois theory it uses is in any case very elementary. And of course the key step is to show that $\vert$Gal$(k/\mathbb{R})\vert = 2^m$ which cannot be done without Sylow.

If anything this application illustrates exactly what a course on group theory should illustrate which is that a lot of analytic-flavoured questions involve insights that are much more adequately generalized in the framework of algebraic structures. (Of course some elementary analytic facts are still required in the proof.)

You can prove the Fundamental Theorem of Algebra using Sylow 2-subgroups. The sketch of the proof is as follows:

1. Take a (WLOG) normal extension $k$ of $\mathbb{R}$ of degree $n$ and a 2-subgroup $H$ of $G=$Gal$(k/\mathbb{R}$). Then $[\text{Fix}(H):\mathbb{R}]=\vert G \vert / \vert H \vert $ and hence since $H$ is a 2-group $n$ must be odd and hence trivial. So $G=H$, i.e. $\vert G \vert = 2^m$
2. Therefore, by taking a subgroup $N$ of $G$ of index $2$ show that Fix$(N)/\mathbb{R}$ is $\mathbb{R}(\theta)$ for a negative square root $\theta$ and hence that Fix$(N)=\mathbb{C}$.
3. Then by a quick argument Gal$(k/\text{Fix}(N))=$Gal$(k/\mathbb{C})=1$ and hence $k=\mathbb{C}$. Hence $\mathbb{C}$ is the only extension of $\mathbb{R}$ and the FTA follows immediately.

That said, students will most likely encounter Sylow p-groups before Galois Theory. But that's not really an argument against providing the proof of FTA as a 'spectacular' application - especially if students are familiar with some basic field theory. The Galois theory it uses is in any case very elementary. And of course the key step is to show that $\vert$Gal$(k/\mathbb{R})\vert = 2^m$ which cannot be done without Sylow.

If anything this application illustrates exactly what a course on group theory should illustrate which is that a lot of analytic-flavoured questions involve insights that are much more adequately generalized in the framework of algebraic structures. (Of course some elementary analytic facts are still required in the proof.)