@above you can actually have $x_1+x_2 = 8$, as well. It seems best to argue from your main equation that $x_3 \geq 3$ is impossible and when $x_3 = 1$ and $x_3=2$ we get two possible different sequences, though amusingly both have $x_7 = 3456$.

We have $x_6 = x_5(x_4 + x_3) = x_4(x_3+x_2)(x_4+x_3) = x_3^2(x_2+x_3)(x_1+x_2)(x_1+x_2+1).$ If we suppose $x_3 \geq 3$, then the right-hand side is at least $3^2 \cdot (3+1) \cdot (1+1) \cdot (1+1+1) = 216$, contradiction. Hence $x_3 = 1$ or $x_3 = 2$. If $x_3 = 1$, then $144 = (x_2 + 1)(x_1+x_2)(x_1+x_2+1)$. For $x_2\geq 4$ the right-hand side is at least $5\cdot 5 \cdot 6 = 150 > 144$, impossible. For $x_2 = 3$ we have $(x_1+3)(x_1+4) = 36$, which has no solutions; if $x_2 = 2$ we have $x_1(x_1 + 5) = 42$ (no solution), and if $x_2 = 1$, then $x_1(x_1 + 3) = 70$, so $x_1 = 7$. Hence $x_1 = 7$, $x_2=x_3 = 1$, thus $x_4 = 8$, $x_5 = 16$, $x_6 = 144$ and $x_7 = 3456$. If $x_3 = 2$, then $36 = (x_2+2)(x_1+x_2)(x_1+x_2+1)$. If $x_2 \geq 2$, the right-hand side is at least $4 \cdot 3 \cdot 4 = 48 > 36$, impossible. Thus $x_2 = 1$, so $(x_1+1)(x_1+2) = 12$, i.e. $x_1 = 2$. Therefore $x_4 = 6$, $x_5 = 18$, $x_6 = 144$ and $x_7 = 3456$.