There's probably a more clever solution.
Case 1: Suppose one of the $x_i$ is $0$: say $x_1$ is $0$. Then,
so that $x_2=x_4$. This implies
This means $x_2=0$ or $x_2=-1$. But $x_3^2=x_1+x_2=x_2$ so $x_2\geq 0$ therefore we infer that $x_2=x_4=0$. From here, it's easy to see that $x_3=x_5=0$ as well.
Case 2: each $x_i$ is nonzero. Label the given equations as (1)-(5). From (1) and (2),
Suppose $x_1>x_3$, then because $x_5^2>0$, we have $x_3>x_4$. In particular, we have $x_1>x_4$, which implies $x_2>x_1$ thanks to equations (4) and (5). So we have $x_2>x_1>x_3>x_4$. Equations (2) and (3) now imply that $x_4>x_5$. But then equations (3) and (4) imply $x_5>x_1$. So you have the impossibility $x_2>x_1>x_3>x_4>x_5>x_1$.
Supposing $x_1<x_3$ will lead to a similar contradiction as in the previous paragraph.
It must be that $x_1=x_3$ so that $x_3=x_4$. But then $x_1=x_4$ implies $x_1=x_2$. In turn, $x_2=x_4$ implies $x_4=x_5$. So $x_1=x_2=x_3=x_4=x_5=x$ for some nonzero $x$. You can now easily see that $x=2$.
Conclusion: 2 solutions $x_i=0$ $\forall i$ and $x_i=2$ $\forall$ $i$.