Base Case $n=1$: if $ax+b=0$ for all values of $x$, then $a=b=0$
\begin{align} ax+b=0\\ (0)+b=0\\ b=0\\\\ ax+(0)=0\\ ax=0\\ a(1)=0\\ a=0 \end{align}
Inductive Step: if $(a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0)\Rightarrow(a_n=a_{n-1}=...=a_0=0)$ for all values of $x$, then $(a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1+a_0=0)\Rightarrow(a_{n+1}=a_n=...=a_0=0)$ is also true.
$$a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1+a_0=0$$ $$a_{n+1}(0)^{n+1}+a_n(0)^n+...+a_1(0)^1+a_0=0$$ $$\Rightarrow a_0=0$$
Now all that is left is to show that all the coefficients of $a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1=0$ are zero.
$$a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1=0$$ $$x(a_{n+1}x^n+a_nx^{n-1}+...a_1)=0$$
Since I'm not interested in the $x=0$ possibility, set the right factor to equal $0$. Since that is in the form of what we are assuming to be true, we've shown that if the theorem works for $n$, it works for $n+1$.
$\therefore a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0\Rightarrow a_n=a_{n-1}=...a_1=a_0=0$, $\forall n\forall x$
Is this proof good enough? I'm a tiny bit unsure about where I just neglected the $x=0$ case. Is the inductive reasoning correct?
Your proof has a gap. The equation $$x(a_{n+1}x^n+a_nx^{n-1}+...+a_1)=0 \tag{1}$$ means $x = 0$ or $a_{n+1}x^n+a_nx^{n-1}+\ldots+a_1 = 0$. This allows to conclude that $$p(x) = a_{n+1}x^n+a_nx^{n-1}+\ldots+a_1 = 0 \text{ for all } x \ne 0 . \tag{2}$$
Induction only works if we know that also $p(0) = 0$. The argument needed here is continuity. All polynomials are continuous, thus $$p(0) =\lim_{x \to 0} p(x) = 0 .$$