Hint: At the bottom of the wikipedia article on the matrix form of the Fibonacci sequence are reported the relations
\begin{align}
{F_m}{F_n} + {F_{m-1}}{F_{n-1}} &= F_{m+n-1},\\
F_{m} F_{n+1} + F_{m-1} F_n &= F_{m+n} .
\end{align}
In particular, with $m=n$,
\begin{align}
F_{2n-1} &= F_n^2 + F_{n-1}^2\\
F_{2n} &= (F_{n-1}+F_{n+1})F_n\\
&= (2F_{n-1}+F_n)F_n .
\end{align}
Using these with $m=k\cdot n$ so that $m+n=(k+1)n$ and induction will lead you to the result.
Full proof:
We want to show that if $m\mid n$ then $F_m\mid F_n$.
Now, $m\mid n$ is equivalent to showing that $n=km\ \ k\in\mathbb{N}$. The induction will be over $k$.
Obviously $k=1$ ($m=n$) is true.
Now let's assume the statement is true for $1\leq k\leq K-1$, let's show it is true for $n=mK$ as well. By the second equality, choosing $n=(K-1)m$ we obtain
\begin{align*}
F_{n}=F_{Km}=F_{m+(K-1)m}&=F_{m}F_{(K-1)m+1}+F_{m-1}F_{(K-1)m}.
\end{align*}
But now by induction hypothesis $F_{(K-1)m}$ is divisible by $F_m$ thus $F_{(K-1)m}=F_m\cdot d$ and hence
\begin{align*}
F_{n}=F_{m}F_{(K-1)m+1}+F_{m-1}F_m\cdot d=F_m\cdot (F_{(K-1)m+1}+F_{m-1}d).
\end{align*}
which shows that $F_m\mid F_n$
