6. Let $\left\{f_{n}\right\}(n \geqslant 0)$ be the Fibonacci sequence, defined as follows:
$$
\begin{array}{l}
f_{0}=1, f_{1}=1, \\
f_{n+1}=f_{n}+f_{n-1}(n=1,2, \cdots) .
\end{array}
$$

Then the solution set of the equation $n f_{n} f_{n+1}=\left(f_{n+2}-1\right)^{2}$ is
$\qquad$ .
6. $\varnothing$.
$$
\begin{array}{l}
\text { Clearly, } f_{n+2}=f_{n+1}+f_{n} \leqslant 2 f_{n+1} \text {. } \\
\text { Then } n f_{n} f_{n+1}=\left(f_{n+2}-1\right)^{2}<f_{n+2}^{2} \\
\leqslant 2 f_{n+1} \cdot 2 f_{n+1} \leqslant 8 f_{n} f_{n+1} \\
\Rightarrow n<8 .
\end{array}
$$

And $f_{0}=1, f_{1}=1, f_{2}=2, f_{3}=3$,
$$
f_{4}=5, f_{5}=8, f_{6}=13, f_{7}=21,
$$

Therefore, when $n \leqslant 7$, $n f_{n} f_{n+1}$ is not a perfect square. Thus, the original equation has no solution.