There is a proof of the fundamental theorem of algebra for $\mathbb{H}$ which has not been cited in the links provided in the comments, and which does not use any topological argument. This concerns polynomials with a central indeterminate $X$ (that is $aX=Xa$ for any quaternion $a$).
The source is Lam's First course in Noncommutative rings.
I sketch the proof here.
Let $D$ be a division ring and let $f=\sum_n d_n X^n\in D[X]$.
We say that $\alpha\in D$ is a right root of $f $ if $f(\alpha):=\sum_n d_n\alpha^n=0$, and a left root if $\sum_n \alpha^n d_n=0$ (note that the latter sum is NOT denoted by $f(\alpha)$.)
First step. Using left/right long division of polynomials, we see that $\alpha $is a left/right root if $f=(X-\alpha)g / f=g (X-\alpha)$, for some $g\in D[X]$. This result generalizes to arbitrary rings (not only division ones).
Second step. Assume that $f=gh$, and let $d\in D$ such that $a:=h(d)\neq 0$. Then $f(d)=g(ada^{-1}))h(d)$ (easy computations). In particular, if $\alpha$ is a right root of $f$, then $\alpha$ is a right root of $h$ or there exists a conjugate of $\alpha$ which is a right root of $g$.
Third step (Main theorem). Let $D=\mathbb{H}$. Then every nonzero polynomial of $D[X]$ has a left/right root in $D$.
For $d=x+yi+zj+tk$, set $d^*=x-yi-zj-tk$.
If $f=\sum_n d_n X^n$ set $f^*=\sum_n d_n^*X^n$. For all $f,g\in D[X]$, one may check that $(fg)^*=g^* f^*$. In particular $(ff^*)^*=f^{**}f^*=ff^*$, and $ff^*\in\mathbb{R}[X]$.
We now prove the theorem by induction on the degree of $f$. For degree $1$, this is obvious.
If $f$ has at least degree $2$, then $ff^*\in\mathbb{R}[X]$ has a root of $\alpha\in \mathbb{R}(i)$ (which is a copy of $\mathbb{C}$ inside $D$) by the classical fundamental theorem of algebra. By step two, either $\alpha$ is a right root of $f^*$ or some conjugate of $\alpha$ is a right root of $f$. In the second case, we are done.
In the first case, $\alpha^*$ is then a left root of $f$, so $f=(X-\alpha^*)g$ for some $g$. Since $1\leq \deg(g)<\deg(f)$, by induction, $g$ has a right root $\beta$. But $\beta$ is also a right root of $f.$ Indeed, we have $g=h(X-\beta)$, and so $f=[(X-\alpha^*)h](X-\beta)$ (note that we cannot use evaluation at $\beta$ to conclude, since it is not a ring morphism).
By induction, we are done. Note that you can prove similarly that any $f$ has a left root.
Concerning octonions : it seems that the partial results are known for polynomials with a noncentral indeterminates. Thus it is possible that the fundamental theorem of algebra is true for octonions if you consider (as I did for quaternions) the indeterminate $X$ to be central.
However, i'm not sure that the proof above maybe easily generalized to octonions because of the lack of associativity.