Subtracting the second equation from the first yields $x_2 - x_3 = x_3^2 - x_2^2 + 6x_4(x_3 - x_2)$, which we can factor as $0 = (x_3 - x_2)(x_3 + x_2 + 1 + 6x_4)$. We see that $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_4 = 0$. Similarly, we also have either $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_1 = 0$. Hence, if $x_2 \neq x_3$, the second equality must hold in both cases; subtracting one from the other, we obtain $x_1 = x_4$. We conclude that either $x_2 = x_3$ or $x_1 = x_4$. Analogously, we get for each permutation $(i, j, k, l)$ of $(1, 2, 3, 4)$ that either $x_i = x_j$ or $x_k = x_l$.
We will prove that at least three of the $x_i$ must be equal. If all four are equal, this is true of course. Otherwise, there are two unequal ones, say $x_1 \neq x_2$ without loss of generality. Then we have $x_3 = x_4$. If also $x_1 = x_3$ holds, then there are three equal elements. Otherwise, we have $x_1 \neq x_3$, hence $x_2 = x_4$ and we also get three equal elements. Up to order, the quadruple $(x_1, x_2, x_3, x_4)$ is thus equal to a quadruple of the shape $(x, x, x, y)$, where we could have that $x = y$.
Substituting this in the equations gives $x+y = 8x^2$ and $2x = x^2 + y^2 + 6xy$. Adding these two equations: $3x + y = 9x^2 + y^2 + 6xy$. The right hand side can be factored as $(3x + y)^2$. Defining $s = 3x + y$, the equation becomes $s = s^2$, from which we get either $s = 0$ or $s = 1$. We have $s = 3x + y = 2x + (x + y) = 2x + 8x^2$. Hence, $8x^2 + 2x = 0$ or $8x^2 + 2x = 1$.
In the first case, we have $x = 0$ or $x = -\frac{1}{4}$. We find $y = 0 - 3x = 0$ and $y = 0 - 3x = \frac{3}{4}$, respectively. In the second case, we get the factorisation $(4x - 1)(2x + 1) = 0$, hence $x = \frac{1}{4}$ or $x = -\frac{1}{2}$. We find $y = 1 - 3x = \frac{1}{4}$ or $y = 1 - 3x = \frac{5}{2}$, respectively.
Altogether, we found the following quadruples: $(0, 0, 0, 0)$, $(-\frac{1}{4}, -\frac{1}{4}, -\frac{1}{4}, \frac{3}{4})$, $(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4})$ and $(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \frac{5}{2})$, and permutations thereof. It is a simple computation to verify that all these quadruples are indeed solutions to the equations. $\square$