Noam Elkies left a very useful comment underneath Qiaochu's Qiaochu Yuan's answer, providing a completely elementary and self-contained solution to my question. For the benefit of anyone interested, I'll spell it out here. (And I'll make this answer community wiki as this is Noam's solution, not mine.)
Let $k$ be an algebraically closed field. For $f \in k[x]$ and $\alpha \in k$, write $\mu(f, \alpha)$ for the multiplicity of $\alpha$ as a root of $f$; that is, $\mu(f,\alpha) = \sup \{n \in \mathbb{N} : (x - \alpha)^n|f(x)\}$. Some very basic facts:
$\mu(f,\alpha) > 0$ iff $f(\alpha) = 0$
$\sum_{\alpha \in k} \mu(f,\alpha) = \sum_{\alpha \in f^{-1}(0)} \mu(f,\alpha) = \deg(f)$ as long as $f \neq 0$
$\mu(f',\alpha) \geq \mu(f,\alpha) - 1$.
A point $a \in k$ is unusual for $f$ iff $\mu(f-a,\alpha) \geq 2$ for all $\alpha \in f^{-1}(a)$.
For nonconstant $f$, it's easy to see that if $a$ is unusual for $f$ then $|f^{-1}(a)| \leq \deg(f)/2$. Indeed, $$ \deg(f) = \deg(f - a) = \sum_{\alpha \in f^{-1}(a)} \mu(f - a,\alpha) \geq 2|f^{-1}(a)|. $$
Theorem Let $f \in k[x]$. If $f' = 0$ then every point of $k$ is unusual for $f$. If $f' \neq 0$ then at most one point of $k$ is unusual for $f$.
Proof The first statement is clear. For the second, suppose that $a$ and $b$ are unusual, with $a \neq b$. We have $$ \begin{aligned} \sum_{\alpha \in f^{-1}(a)} \mu(f', \alpha)& = \sum_{\alpha \in f^{-1}(a)} \mu((f - a)', \alpha)\\ &\geq \sum_{\alpha \in f^{-1}(a)} \bigl[\mu((f - a), \alpha) - 1\bigr]\\ & = \deg(f-a) - |f^{-1}(a)|\\ & \geq \frac{1}{2}\deg(f). \end{aligned} $$ The same goes for $b$. But $f^{-1}(a) \cap f^{-1}(b) = \emptyset$ and $f' \neq 0$, so $$ \begin{aligned} \deg(f') & = \sum_{\gamma \in k} \mu(f', \gamma)\\ & \geq \sum_{\alpha\in f^{-1}(a)} \mu(f', \alpha) + \sum_{\beta\in f^{-1}(b)} \mu(f',\beta)\\ & \geq \frac{1}{2}\deg(f) + \frac{1}{2}\deg(f) = \deg(f), \end{aligned} $$ giving $\deg(f') \geq \deg(f)$, a contradiction.
