Is there a Fibonacci number whose square is another Fibonacci?

Let $n>1$. Then $F_n\mid F_m$ if and only if $n\mid m$. (Assuming we start at $F_0=0,F_1=1$.)

This is because we get, inductively, $F_{n+k}\equiv F_{n+1}F_k\pmod{F_n}$, and $F_{n+1}$ and $F_{n}$ are relatively prime.

But you can use Demoivre's formula to show that for $n\geq 2$ that $F_n^2<F_{2n}$.

Specifically, with $\phi=\frac{1+\sqrt{5}}{2}$ and $\rho = \frac{-1}{\phi}=\frac{1-\sqrt{5}}{2}$ we have that:

$$F_{n}=\frac 1{\sqrt{5}}\left(\phi^n-\rho^n\right)$$

And $F_{2n}=F_n\left(\phi^n+\rho^n\right)$. So you need to snow that $\phi^{n}+\rho^n >\frac{1}{\sqrt{5}}(\phi^n-\rho^n)$, or $$(\sqrt{5}-1)\phi^n>-(\sqrt{5}+1)\rho^n$$

a non rigorous but maybe easy to understand proof:

Assume such a number $x$ exists, $x > 1, a < x,$ then the sequence goes with, $a, x, x+a, 2x+a, 3x+2a, 5x+3a, ...$

notice the coefficients of the linear expression are in Fibonacci sequence. So that

$ax + (x-a)a,$
$x^2 + a^2 $

are next to each other in the sequence.

But $ax + (x-a)a = x^2 - (x-a)^2 < x^2,$ and the next one $> x^2$. so $x^2$ is not in the sequence.

Working off of Ryan Y's approach:

Note that we have $F_{2n-1} = F_{n-1}^2+F_n^2$; thus, we know that $F_{2n-1}\gt F_n^2$ as long as $F_{n-1}^2\gt 0$, which will happen for all $n\gt 1$.

But we also have $F_{2n-2} = 2F_nF_{n-1}-F_{n-1}^2$, and so $F_n^2\gt F_{2n-2}$ as long as $F_n^2-F_{2n-2}\gt 0$ — or, in other words, $F_n^2-2F_nF_{n-1}+F_{n-1}^2\gt 0$, or $(F_n-F_{n-1})^2\gt 0$. And so this holds whenever $F_n\gt F_{n-1}$, which will hold as soon as $n\gt 2$.

Thus, we have explicitly that $F_{2n-2}\lt F_n^2\lt F_{2n-1}$ for all $n\gt 2$.

(These doubling identities are classical, and there are many ways to derive them; I prefer the matrix-based approach seen in this answer because it's very general and it almost immediately generalizes even further to other linear recurrence sequences.)