Let $F_{u_{p^1}}$, denote the least positive Fibonacci number divisible by the prime $p$.

Let $F_{u_{p^2}}$, denote the least positive Fibonacci number divisible by $p^2$.

Let $\prod_{1}^{n\ge 1}$ be a quotient of primitive(characteristic) prime factor(s), ie factors that have not occurred in any earlier Fibonacci numbers.
This may be represented below as an empty product of one, if the quotient divides $F_{p^{e}_{1}},\ e>1$.

$\left \lbrace {n \mid m}\ \text{iff}\ {F_n \mid F_m} \right \rbrace\, n \ge 3 $ David Wells 1986, p.65

$p^2 \mid F_{u_{p^1}}\ \text{iff}\ F_{p^2} \mid F_{F_{u_{p^1}}}$

If an integer $i$, has prime factorization, $p_{1}^{e_{1}}\cdot\ p_{2}^{e_{2}}...p_{n}^{e_{n}}$ then the entry point of $i$ equals, $\left \lbrace u_{i} = \operatorname{lcm}[{ u_{p_{1}^{e_{1}}}, u_{p_{2}^{e_{2}}}, ...u_{p_{n}^{e_{n}}}}] \right \rbrace$.Marc Renault 1996 p.19 theorem 3.3 http//webspace.ship.edu/msrenault/fibonacci/FibThesis.pdf

$\text{If}\ i=F_{F_{u_{p^1}}}\ \text{then, }\ F_{p^2} \nmid\! F_{F_{u_{p^1}}}\text{ because } \left \lbrace u_{(F_{F_{u_{p^1}}})}\neq u_{(F_{F_{u_{p^2}}})} \right \rbrace$.

An abstract example of a Fibonacci-Wieferich prime p, where $p_{1}=p,\ e_{1}\ge 2$.
If $i=F_{F_{u_{p^1}}}=(F_{p_{1}^2}\ \cdot\ F_{p_{2}^{e_{2}}}\ \cdot\ ...\ F_{p_{n}^{e_{n}}}\ \cdot\ \prod{})$ then, can the entry point of $i$ still equal the same index, ie $u_{(F_{F_{u_{p^1}}})}=F_{u_{p^1}}$?
Answer No.
Proof
$u_i=\operatorname{lcm}[p_{1}^{2},\ p_{2}^{e_{2}},\ ...\ p_{n}^{e_{n}},\ F_{u_{p^1}}]=\operatorname{lcm}[p_{1}^{2},\ p_{2}^{e_{2}},\ ...\ p_{n}^{e_{n}}]\neq F_{u_{p^1}}=F_{p\cdot u_{p^1}}$
Since any extra powers of $p_{1}$, produce a later entry point, thus $F_{p^2}$ does not appear as a factor, in this configuration, until a later Fibonacci number, ie $F_{F_{p\cdot u_{p^1}}}=F_{F_{u_{p^2}}}$.
While the factor, $j=(F_{p_{1}^1}\ \cdot\ F_{p_{2}^{e_{2}}}\ \cdot\ ...\ F_{p_{n}^{e_{n}}}\ \cdot\ \prod{})$ always has an entry point of, $u_j=\operatorname{lcm}[p_{1}^1,\ p_{2}^{e_{2}},\ ...\ p_{n}^{e_{n}},\ F_{u_{p^1}}]=\operatorname{lcm}[p_{1}^1,\ p_{2}^{e_{2}},\ ...\ p_{n}^{e_{n}}]=F_{u_{p^1}}$.