Claim i. $f(x)$ is injective.
Proof. Assume there exist two natural numbers $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$ and $x_1 \neq x_2$. Now choose $y$ and $z$ such that $y + z = t^2 - x_1$ then we have
$$f(y) + f(z) + f(x_1) = a^2$$
for some natural number $a$. Therefore,
$$f(y) + f(z) + f(x_1) = f(y) + f(z) + f(x_2) = a^2,$$
so $y + z + x_2$ must be square of some natural number $s$. Hence
$$|x_1 - x_2| = |t^2 - s^2| \neq 0$$
and we have
$$|t^2 - s^2| \geq 2|t| - 1.$$
If we choose $t$ large enough it gives us a contradiction.

Claim ii. For any $x, y \in \mathbb{N}$ we have $f(x + 1) - f(x) = f(y + 1) - f(y)$.
Proof. Assume some arbitrary $x_0$ and $y_0$ with $x_0 \geq 2$, and choose $z = t^2 - x_0 - y_0$ for some large $t$. Then
$$f(x_0) + f(y_0) + f(z) = a^2,$$
and
$$f(x_0 - 1) + f(y_0 + 1) + f(z) = b^2.$$
So we have
$$|f(x_0) + f(y_0) - f(x_0 - 1) + f(y_0 + 1)| = |a^2 - b^2|.$$
Assume that $f(x_0) + f(y_0) - f(x_0 - 1) + f(y_0 + 1) \neq 0$. If we choose $z$ large enough, $f(z)$ is a large number too (it's trivial because of the injectivity). But we know
$$|f(x_0) + f(y_0) - f(x_0 - 1) + f(y_0 + 1)| = |a^2 - b^2| \geq 2|a| - 1.$$
That means $f(z)$ has an upper bound. Contradiction.

Claim ii implies that $f(x)$ is a linear function so $f(x) = a x + b$ and
$$a(x + y + z) + 3b,$$
becomes a perfect-square if and only if
$$x + y + z,$$
does. That means $a t^2 + 3b$ is a perfect-square for any $t \in \mathbb{N}$. In particular $a(a t^2 + 3b) + 3b = t'^2$ hence
$$3ba + a = t'^2 - t^2,$$
which is a contradiction unless $3ba + 3b = 0$ (because the left hand side is either 0 or very big for big $t$). Now obviously $a > 0$ and we get $b = 0$, $a = n^2$.