I'm trying to prove this for all $n \geq 1$. Using the recursive formula, I ended up with this: $F_{-(n+1)} = F_{-n} - F_{-(n+2)}$.
If the formula holds for $n$ and $n+2$, I can eventually turn the above equation into the desired result $F_{-(n+1)} = (-1)^{n+2}F_{n+1}$.
So, if I show that the formula holds for $n = 1$ and $n = 3$ (which it does), is that enough to assume that formula holds for $n$ and $n+2$ for some integer n? Or do I need to also show that the formula holds for $n=2$?
Your approach won’t work, I’m afraid: it requires you to know the truth of the result for $n+3$ in order to get it for $n+2$. If you know it initially only for $n=1$ and $n=3$, you can get it for $n=2$ but not for anything else: to get it for $n=4$, you would need to know it for $n=5$ already, to get it for $n=5$ you would need to know it for $n=6$, and so on.
You need to rewrite the recurrence so that the number with the smallest (i.e., most negative) subscript is on the left: $F_{-n}=F_{-(n+1)}+F_{-(n+2)}$, so
$$F_{-(n+2)}=F_{-n}-F_{-(n+1)}\;.\tag{1}$$
Now take $n=1$ and $n=2$ as your base cases: by actual computation we have
$$F_{-1}=F_1-F_0=1-0=1=(-1)^2F_1$$
and $$F_{-2}=F_0-F_{-1}=0-1=-1=(-1)^3F_2\;.$$
I’ll leave the induction step to you; you should use $(1)$ to carry it out.