Let's assume that \(m \lt n\). As shown here, this means:
$$ F_0 F_1 F_2 \cdots F_m \cdots F_{n-1} + 2 = F_n$$
Rearranging:
$$\begin{gather} 2 = F_n - F_0 F_1 F_2 \cdots F_m \cdots F_{n-1} \\ F_n - F_m k = 2 \end{gather}$$
Any divisor that divides \(F_n\) and \(F_m\) should also be able to divide any linear combination of \(F_n\) and \(F_m\). Let \(d\) divide \(F_n - F_m k\):
$$ d | F_n - F_m k \implies d|2 $$
This means \(d=2\) or \(d=1\). Since both \(F_n\) and \(F_m\) are odd, then \(d\) cannot be 2. Therefore \(d=1\). Since the smallest linear combination of two numbers is their gcd, then \((F_n, F_m) = 1\).