This is not a serious answer, but one can "prove" the fundamental theorem of algebra by applying the spectral theorem to the matrix

$$ A := \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\\ 0 & 0 & 1 & \ldots & 0 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ -a_0 & -a_1 & -a_2 & \ldots & -a_{n-1} \end{pmatrix} $$

to locate an eigenvalue $\lambda$, which is then a root of the monic polynomial $z^n + a_{n-1} z^{n-1} + \ldots + a_0$.

Of course, this argument is usually circular, because most of the standard proofs of the spectral theorem for matrices requires the fundamental theorem of algebra (either by explicitly citing that theorem, or implicitly, by borrowing one of the proofs given here, e.g. by applying Liouville's theorem to the resolvent $(A-zI)^{-1}$) in the first place...

However, one could imagine a weird proof of the spectral theorem that somehow avoids the fundamental theorem and would thus give a non-circular proof of that theorem. I thought about proceeding by showing that the set of diagonalisable matrices is a generic subset of the set of all matrices, but I realised that in order to have enough algebraic geometry to talk about "generic", I need to know the ambient field is algebraically closed, which of course is precisely the fundamental theorem of algebra.

Perhaps it is best to view the above arguments not as proofs of the fundamental theorem of algebra, but rather as "consistency checks" that show that this result is compatible with the basic theory of other mathematical subjects, such as linear algebra and algebraic geometry.

This is not a serious answer, but one can "prove" the fundamental theorem of algebra by applying the spectral theorem to the matrix

$$ A := \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\\ 0 & 0 & 1 & \ldots & 0 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ -a_0 & -a_1 & -a_2 & \ldots & -a_{n-1} \end{pmatrix} $$

to locate an eigenvalue $\lambda$, which is then a root of the monic polynomial $z^n + a_{n-1} z^{n-1} + \ldots + a_0$.

Of course, this argument is usually circular, because most of the standard proofs of the spectral theorem for matrices requires the fundamental theorem of algebra (either by explicitly citing that theorem, or implicitly, by borrowing one of the proofs given here, e.g. by applying Liouville's theorem to the resolvent $(A-zI)^{-1}$) in the first place...

However, one could imagine a weird proof of the spectral theorem that somehow avoids the fundamental theorem and would thus give a non-circular proof of that theorem. I thought about proceeding by showing that the set of diagonalisable matrices is a generic subset of the set of all matrices, but I realised that in order to have enough algebraic geometry to talk about "generic", I need to know the ambient field is algebraically closed, which of course is precisely the fundamental theorem of algebra. Deducing the spectral theorem for matrices from the spectral theory of more general objects, such as elements of a C^* algebra, doesn't work either, for much the same reason.

Perhaps it is best to view the above arguments not as proofs of the fundamental theorem of algebra, but rather as "consistency checks" that show that this result is compatible with the basic theory of other mathematical subjects, such as linear algebra and algebraic geometry.