For simplicity write $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc. We first show
$$f^2(x) = x \quad \text{for all } x > 0$$
in two different ways.

Method 1, using injectivity. Suppose $f(z_1) = f(z_2)$. Replacing $z$ by $z_1$ or by $z_2$ leads to the same LHS of (2), hence the RHS must agree in both cases as well, i.e. $z_1 + f(y) + f^2(x) = z_2 + f(y) + f^2(x)$ which implies $z_1 = z_2$, hence $f$ must be injective.

We make the RHS of (2) equal to $z + f^2(y) + f^2(x)$ in two different ways, in order to benefit from the injectivity of $f$. Replacing $y$ by $f(y)$ gives $f(x + f^2(y) + f^2(z)) = z + f^2(y) + f^2(x)$. Replacing $x$ by $y$ and $y$ by $f(x)$ gives $f(y + f^2(x) + f^2(z)) = z + f^2(x) + f^2(y)$. Because the RHS is the same in both equations, and $f$ is injective, it follows that $x + f^2(y) + f^2(z) = y + f^2(x) + f^2(z)$, i.e. $f^2(x) - x = f^2(y) - y$ for all positive $x, y$.

This means that $f^2(x) - x = c$ is a constant that does not depend on $x$. Setting $t = 1 + f(1) + f^2(1)$, equation (2) with $x = y = z = 1$ implies $f(t) = t$, hence $f^2(t) = t$ and $f^2(t) - t = 0$. Therefore $c = 0$ and we have shown $f^2(x) = x$ for all $x > 0$.

Method 2, using fixed points. (Tianci Yan) Putting $x = z$ in (2) gives
$$f(x + f(y) + f(f(x))) = x + f(y) + f(f(x))$$
and this means that $x + f(y) + f(f(x))$ is a fixed point for all $x, y > 0$.

Let $a$ be any fixed point of $f$. Putting $x = y = a$ in (2) gives
$$f(2a + f(f(z))) = z + 2a, \text{ for } z > 0.$$

Putting $y = z = a$ in (2) gives
$$f(x + 2a) = 2a + f(f(x)), \text{ for } x > 0.$$
Applying $f$ to both sides of the previous equation and using the earlier result gives
$$f(f(x + 2a)) = x + 2a, \text{ for } x > 0.$$
Replacing $x$ by $x + 2a$ in the previous equation then gives
$$f(x + 4a) = 2a + f(f(x + 2a)) = x + 4a, \text{ for } x > 0.$$
So $f(w) = w$, for all integers $w > 4a$. Pick such an integer $w > 4a$ and put $y = z = w$ in the original equation (2) to get
$$f(x + f(w) + f(f(w))) = w + f(w) + f(f(x))$$
which can be simplified to $x + 2w = f(x + 2w) = 2w + f(f(x))$ for $x > 0$.
This shows that $f(f(x)) = x$ for all $x > 0$, which is $f^2(x) = x$.

Having established $f^2(x) = x$, we can complete the solution of the problem as follows. Given an integer $n \ge 3$, we set $x = n - 2$, $y = f(1)$ and $z = 1$ in (2). With the aid of $f^2(x) = x$ this becomes $f(n) = n$ for all $n \ge 3$. Because of $f^2(x) = x$, we then must have $f(1) < 3$ and $f(2) < 3$. Again from $f^2(x) = x$ we see that $f(1) = 2$ implies $f(2) = 1$ and vice versa. This leaves us with two possible solutions:
$$\begin{aligned} f_1(n) &= n \text{ for } n \ge 0 \\ f_2(n) &= n \text{ for } n \ge 3, \quad f_2(1) = 2, \quad f_2(2) = 1. \end{aligned}$$
Clearly, $f_1$ satisfies (2). To verify the functional equation for $f_2$ we first note that $x + f_2(y) + f_2^2(z) \ge 3$, hence (2) is satisfied if and only if
$$x + f_2(y) + f_2^2(z) = z + f_2(y) + f_2^2(x),$$
i.e. $f_2^2(x) - x = f_2^2(z) - z$ for all $x, z \in \mathbb{Z}_+$. Clearly $f_2^2(z) - z = 0$ for $z \ge 3$. Since $f_2(f_2(1)) = f_2(2) = 1$ and $f_2(f_2(2)) = f_2(1) = 2$ we see that indeed $f_2^2(x) - x = 0$ for all $x \in \mathbb{Z}_+$. This shows that $f_2$ is a solution as well.