A result of the type you seek follows easily from [Carmichael's theorem][1], that if $m > 12$, then there is a prime $p$ that divides $F_{m}$, but does not divide $F_{k}$ for $k < m$. 

Suppose $C_{n} = \binom{2n}{n}/(n+1)$ and we assume that $C_{n} = F_{m}$. All the prime factors of the left hand side are $\leq 2n$, while by Carmichael's theorem there is a prime $p | F_{m}$ that does not divide any earlier Fibonacci number. However, since for $p > 5$, $p$ divides either $F_{p-1}$ or $F_{p+1}$ we have that $m \leq p+1 \leq 2n$. However, from the asymptotic size of $C_{n}$ and $F_{m}$, we must have that $m \approx n^{\log(4)/\log(\phi)}$, and this contradicts the inequality $m \leq 2n$.

  [1]: http://en.wikipedia.org/wiki/Carmichael%27s_theorem