I was playing around with Fibonacci numbers divisible by their indexes (i.e. the $12$th Fibonacci number, $144$, is divisible by $12$) and found that this works when the index is a power of $5$ or $12$. For example:
| $k$ | $N = 5^k$ | $N$th Fibonacci number |
|---|---|---|
| $0$ | $1$ | $1 = 1 * 1$ |
| $1$ | $5$ | $5 = 5 * 1$ |
| $2$ | $25$ | $75025 = 25 * 3001$ |
| $3$ | $125$ | $59425114757512643212875125 = 125 * 475400918060101145703001$ |
and...
| $k$ | $N = 12^k$ | $N$th Fibonacci number |
|---|---|---|
| $0$ | $1$ | $1 = 1 * 1$ |
| $1$ | $12$ | $144 = 12 * 12$ |
| $2$ | $144$ | $555565404224292694404015791808 = 144* 3859669476841564088624933206$ |
So far, I found that this holds true up to $5^8 = 390625$ and $12^5 = 248832$, the largest powers of $5$ and $12$ less than $1$ million. Is it provable that this will work for all powers of $5$ and/or $12$, or is there a counter-example?
Edit: while the old answer has some interest, it’s very laconic, and, although more easy to generalize (it straightforwardly implies the analog of Claims 1 and 2 for any prime $p \neq 5$), needlessly sophisticated for the problem at hand.
Let $a=\frac{1+\sqrt{5}}{2},b=\frac{1-\sqrt{5}}{2}$. It’s well-known that $F_n=\frac{a^n-b^n}{\sqrt{5}}$, that $a+b=1$, $ab=-1$, and that for all $n \geq 1$, $a^n+b^n$ is an integer.
Claim 1: if $3\mid F_n$, then $3F_n\mid F_{3n}$.
Proof: $\frac{F_{3n}}{F_n}=a^{2n}+b^{2n}+(ab)^n=(a^n-b^n)^2+3(ab)^n=5F_n^2+3(-1)^n$.
Claim 2: if $2\mid F_n$, then $2F_n\mid F_{2n}$.
Proof: $\frac{F_{2n}}{F_n}=a^n+b^n$ is an integer. Note that $(a^n+b^n)^2=(a^n-b^n)^2+4(ab)^n=5F_n^2+4(-1)^n$ is even, so $\frac{F_{2n}}{F_n}$ is even.
Claim 3: let $n\mid m$, then $F_n\mid F_m$.
Proof: as said in the comments, this is well-known, but I include a proof for the sake of completeness. Write $m=pn$.
If $p=2l+1$, then $\frac{F_m}{F_n}=(ab)^{nl}+\sum_{k=0}^{l-1}{(ab)^{nk}(a^{n(p-1-2k)}+b^{n(p-1-2k)})}$ is an integer. On the other hand, if $p=2$, then $\frac{F_m}{F_n}=a^n+b^n$ is an integer. Then use induction on the number of prime factors of $p$.
Now, note that $3\mid F_4$. Thus, by Claims 1 and 3, for all $k,r \geq 1$, $3^{r+1}\mid F_{4k\cdot 3^r}$.
Since $2\mid F_3$, by Claims 2 and 3, for all $k,r \geq 1$, $2^{r+1}\mid F_{3k \cdot 2^r}$.
It follows in particular that for any $k,r \geq 1$, $12^{r+1}\mid F_{k \cdot 12^r}$.
Old answer:
Let $p=2$ or $3$, and $O$ the ring of integers of the unique extension of $\mathbb{Z}_p$ (the $p$-adic integers) unramified of degree two.
Note that the identity $F_n=\frac{\phi^n-\overline{\phi}^n}{\sqrt{5}}$ also holds (meaningfully) in $O$, and, for any $k \geq 0$, $p^k$ divides $F_n$ iff $p^k\mid \phi^n-\overline{\phi}^n$ iff $p^k\mid -(\phi\overline{\phi})^n+\phi^{2n}$ iff $p^k\mid\phi^{2n}-(-1)^n$.
Now take $p=3$, $n=4q \cdot 3^r$, and note that $\phi^4=3\phi+2\equiv -1 \pmod{3}$, and $\phi^8=(3\phi+2)^2 \not\equiv 1\pmod{9}$.
By (a suitable extension with an identical proof of) the LTE lemma, we see that $v_3(F_n)=v_3(\phi^{8q\cdot 3^r}-1)=v_3(\phi^8-1)+v_3(3^rq) \geq {r+1}$.
Take $p=2$, $n=3q\cdot 2^r$ (with an odd $q$ and $r\geq 1$), then $\phi^6=(2\phi+1)^2=8\phi+5$, so by “LTE” again (since $4\mid \phi^6-1$), $v_2(F_n)=v_2((\phi^6)^{2^rq}-1)=v_2(\phi^6-1)+r=2+r$.
In particular, one easily sees that $12^{r+1}\mid F_{k \cdot 12^r}$ for all $r,k\geq 1$.