For any integer $k \ge 3$ and any $x \ge 0$ we have
$$2x^k \ge [(k-1)x - (k-2)](x^2 + 1). \quad (0)$$
To see this, observe that, by AM-GM inequality,
$$x^k + x^k + \underbrace{x + x + \dots + x}_{(k-3)\text{ times}} \ge (k-1)x^3,$$
and add it to
$$(k-2)(x^2 - 2x + 1) \ge 0.$$
Note that we have equality if and only if $x=1$.
Therefore, for $x$, $y$, $z$ satisfying the system of equations, we have
$$\begin{align*} 2y(x^2 + 1) - (z^2 + 1) &\ge (2x-1)(x^2 + 1), \\ 3z(y^2 + 1) - 2(x^2 + 1) &\ge (3y-2)(y^2 + 1), \\ 4x(z^2 + 1) - 3(y^2 + 1) &\ge (4z-3)(z^2 + 1), \end{align*}$$
or
$$\begin{equation} \begin{aligned} & 2(y-x)(x^2+1) + (x-z)(x+z) \ge 0, \\ & 3(z-y)(y^2+1) + 2(y-x)(y+x) \ge 0, \\ & 4(x-z)(z^2+1) + 3(z-y)(z+y) \ge 0. \end{aligned} \tag{1} \end{equation}$$
Now suppose $x \ge \max\{y, z\}$. Then from the second inequality of (1) we infer that $y \le z$ and
$$\begin{align*} & 2(y-x)(x^2+1) + (x-z)(x+z) \le (z-x)(2(x^2+1)-(x+z)) \\ & \le (z-x)(2x^2-2x+2) \le 0, \end{align*}$$
which, by the first inequality in (1), implies $x = y = z$.

If $y \ge \max\{x, z\}$, then $z \le x$ by the third inequality in (1) and
$$3(z - y)(y^2 + 1) + 2(y - x)(y + x) \le (x - y)(3(y^2 + 1) - 2y - 2x) \\ \le (x - y)(3y^2 - 4y + 3) \le 0,$$
hence, by the second inequality in (1), $x = y = z$.
Finally, if $z \ge \max\{x, y\}$, then $x \le y$ by the first estimate in (1) and, as previously,
$$4(x - z)(z^2 + 1) + 3(z - y)(z + y) \le (y - z)(4z^2 - 6z + 4) \le 0,$$
which again implies $x = y = z$. Thus we have equality in (0) and hence $x = y = z = 1$. We easily check that this is the solution to the system.