Given a polynomial f, can there be more than one constant c such that every root of f(x)-c is repeated?

This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero).

We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and $f, h$ have all of their roots repeated. If $g$ is nonzero, $f, h$ must be relatively prime. Letting $d = \deg f$, it follows that $fgh$ has at most $d$ roots, but by Mason-Stothers $fgh$ must have at least $d+1$ roots; contradiction.

Encouraged by Pierre-Yves Gaillard here is a quick solution to the original problem. It is based on Tom Leinster's answer which in turn is an elaboration of Noam Elkies's comment to Qiaochu Yuan's answer.

Assume $a\neq b$ are unusual for $f(x)$. Then $f'(x)^2$ is divisible by $f(x)-a$ and $f(x)-b$, hence also by their product. This would show $\deg(f')\geq\deg(f)$, a contradiction.

Extend $f$ to a regular morphism $\bar f:\mathbb P^1\to \mathbb P^1$ and write down the Hurwitz formula: $$K_{\mathbb P^1}\sim \bar f^*K_{\mathbb P^1} +R$$ where $R$ is the ramification divisor. Since $f$ is a polynomial, $\bar f$ is completely ramified at $\infty$, so $R$ contains that point with multiplicity $d-1$. Therefore, by the above equivalence the rest of $R$ has degree $d-1$. Now for a point on the target (your "$c$") for which all the points in the preimage are multiple, the degree of the ramification divisor above this point has to be at least $d-\frac d2=\frac d2$ (it's the degree of the map minus the number of points). If you had two such points their combined degree would be at least $d$ contradicting the previous observation that it should be at most $d-1$.