We can in fact show a stronger statement with some algebraic number theory: If $p>5$ is prime then $p|F_{p\pm 1}$ for some choice of $+$ or $-$.

Suppose $\left(\frac{5}{p}\right)=1$. In this case, $p$ splits in $\mathbf{Z}\left[\frac{1+\sqrt{5}}{2}\right]=\mathbf{Z}[\varphi]$. Thus, we can write $p=\pm\pi\bar\pi$, where $\pi$ and $\bar\pi$ are conjugate primes in $\mathbf{Z}[\varphi]$ that do not differ by a unit. Write $\pi=x+y\varphi$, so $x+y\varphi\equiv 0\pmod{\pi}$. Now, if $p|y$, then $\pi|y,x$, contradiction, so $p\nmid y$. Thus, $y$ has an inverse modulo $p$, say $y'$. Then we have $\pi|p|yy'-1$, so $\varphi\equiv -xy'\pmod{\pi}$. Summarizing, $\varphi\equiv k\pmod{\pi}$ for some integer $k\not\equiv 0\pmod{p}$. By FLT, $k^{p-1}\equiv 1\pmod{p}$, so $\varphi^{p-1}\equiv k^{p-1}\pmod{\pi}$. Thus $\varphi^{p-1}\equiv 1\pmod{\pi}$. Similarly, we see that $\bar\varphi^{p-1}\equiv 1\pmod{\pi}$, so $F_{p-1}\sqrt{5}\equiv 0\pmod{\pi}$. Since $p$ and $5$ are necessarily relatively prime, $\pi|F_{p-1}$, and $\bar\pi|F_{p-1}$. Hence $\pi\bar\pi = p|F_{p-1}$ in this case.

Now, suppose $\left(\frac{5}{p}\right)=-1$. We have $5^{(p-1)/2}\equiv -1\pmod{p}$, by Euler's Criterion. Now, applying the Binomial-theorem to Binet's formula yields several terms containing $\binom{p+1}{k}\equiv 0\pmod{p}$. After reducing modulo $p$ we will be left with $\dfrac{\sqrt{5}^{p+1}+1}{2^t}$ for some $t$, which is also divisible by $p$ (by working in $\mathbf{Z}[\varphi]$), so $p|F_{p+1}$ in this case.

Note: This is a proof of the above post by **01000100**, which asserts that $p|F_{p-\left(\frac{p}{5}\right)}$