Let us find a function $f$ such that the conditions given in the problem statement are satisfied, starting from given hints and systematically addressing each part of the problem.

First, we analyze the condition $f^2(1) + f(1) \mid 4$ and $f^2(1) + f(1) > 1$. Since divisors of 4 greater than 1 are 2 and 4, we can set:

1. If $f^2(1) + f(1) = 2$:
$$f^2(1) + f(1) = 2 \quad \Rightarrow \quad f(1)(f(1) + 1) = 2$$
This equation has no integer solution for $f(1)$.

2. If $f^2(1) + f(1) = 4$:
$$f^2(1) + f(1) = 4 \quad \Rightarrow \quad f(1)^2 + f(1) - 4 = 0$$
Solving this quadratic equation using the quadratic formula:
$$f(1) = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$$
Again, this does not yield integer results. However, testing practical small values give $f(1) = 1$ satisfies as:
$$f^2(1) + f(1) = 1^2 + 1 = 2$$

With $f(1) = 1$, we proceed by considering that for each prime $p$, the number $1 + f(p-1) \mid p^2$ and $1 + f(p-1) > 1$. Thus, $1 + f(p-1) = p$ or $p^2$.

Explore the case where $1 + f(p-1) = p^2$:
$$1 + f(p-1) = p^2 \quad \Rightarrow \quad f(p-1) = p^2 - 1$$
Then,
$$f^2(p-1) + f(1) = p^4 - 2p^2 + 2$$
The expression divides $((p-1)^2 + 1)^2 = p^4 - 4p^3 + 8p^2 - 8p + 4$, but verifying,
$$p^4 - 4p^3 + 8p^2 - 8p + 4 < 2(p^4 - 2p^2 + 2)$$
This leads to the conclusion that $p^4 - 4p^3 + 8p^2 - 8p + 4 = p^4 - 2p^2 + 2$ equating gives:
$$2p^3 - 5p^2 + 4p - 1 = 0$$
Since this equation is impossible for integer $p \geq 2$, as $p \mid 1$ is absurd, thus, for all prime $p$,
$$1 + f(p-1) = p \quad \Rightarrow \quad f(p-1) = p - 1$$

Finally, for all integers $n$ and primes $p$, conclude from:
$$f(n) + (p-1)^2 \mid ((p-1)^2 + n)^2$$
Which implies:
$$\frac{(f(n) - n)^2}{f(n) + (p-1)^2} \text{ is an integer}$$
Choosing sufficiently large $p$, the fraction’s value becomes less than 1, thus:
$$f(n) = n$$

Thus, the function satisfy the problem's conditions, confirming:
$$\boxed{f(n) = n}$$