The desired condition is that $x_{n}=5 x_{n+1}$ or $x_{n+1}=5 x_{n}$. Note that for any prime $p$, we have $\nu_{p}(M(a, b))=\left|\nu_{p}(a)-\nu_{p}(b)\right|$. Furthermore, $\nu_{p}(M(a, b)) \equiv \nu_{p}(a)+\nu_{p}(b) \bmod 2$. So, we have that $$\nu_{p}\left(x_{n}\right) \equiv \nu_{p}(1)+\nu_{p}(2)+\cdots+\nu_{p}(n) \bmod 2$$ Subtracting gives that $\nu_{p}\left(x_{n+1}\right)-\nu_{p}\left(x_{n}\right) \equiv \nu_{p}(n+1) \bmod 2$. In particular, for $p \neq 5, \nu_{p}(n+1)$ must be even, and $\nu_{5}(n+1)$ must be odd. So $n+1$ must be a 5 times a perfect square. There are $\left\lfloor\sqrt{\frac{2021}{5}}\right\rfloor=20$ such values of $n$ in the interval [2, 2021]. Now we show that it is sufficient for $n+1$ to be 5 times a perfect square. The main claim is that if $B>0$ and a sequence $a_{1}, a_{2}, \ldots, a_{B}$ of nonnegative real numbers satisfies $a_{n} \leq B+\sum_{i<n} a_{i}$ for all $1 \leq n \leq N$, then $$\left|a_{1}-\right| a_{2}-|\cdots-| a_{N-1}-a_{N}|| \cdots|| \leq B$$ This can be proved by a straightforward induction on $N$. We then apply this claim, with $B=1$, to the sequence $a_{i}=\nu_{p}(i)$; it is easy to verify that this sequence satisfies the condition. This gives $$\nu_{p}\left(x_{n}\right)=\left|\nu_{p}(1)-\right| \nu_{p}(2)-|\cdots-| \nu_{p}(n-1)-\nu_{p}(n)|| \cdots|| \leq 1$$ so $\nu_{p}\left(x_{n}\right)$ must be equal to $\left(\nu_{p}(1)+\cdots+\nu_{p}(n)\right) \bmod 2$. Now suppose $n+1=5 k^{2}$ for some $k$; then $\nu_{p}(n+1) \equiv 0 \bmod 2$ for $p \neq 5$ and $\nu_{5}(n+1) \equiv 1 \bmod 2$. Therefore $\nu_{p}\left(x_{n+1}\right)=\nu_{p}\left(x_{n}\right)$ for $p \neq 5$, and $\nu_{5}\left(x_{n+1}\right)=\left(\nu_{5}\left(x_{n}\right)+1\right) \bmod 2$, and this implies $x_{n+1} / x_{n} \in\{1 / 5,5\}$ as we wanted.