First, it is clear that all functions of the form $f(n)=n+c$ with a constant nonnegative integer $c$ satisfy the problem conditions since $(f(m)+n)(f(n)+m)=(n+m+c)^{2}$ is a square.

We are left to prove that there are no other functions. We start with the following Lemma. Suppose that $p \mid f(k)-f(\ell)$ for some prime $p$ and positive integers $k, \ell$. Then $p \mid k-\ell$.

Proof. Suppose first that $p^{2} \mid f(k)-f(\ell)$, so $f(\ell)=f(k)+p^{2} a$ for some integer $a$. Take some positive integer $D>\max \{f(k), f(\ell)\}$ which is not divisible by $p$ and set $n=p D-f(k)$. Then the positive numbers $n+f(k)=p D$ and $n+f(\ell)=p D+(f(\ell)-f(k))=p(D+p a)$ are both divisible by $p$ but not by $p^{2}$. Now, applying the problem conditions, we get that both the numbers $(f(k)+n)(f(n)+k)$ and $(f(\ell)+n)(f(n)+\ell)$ are squares divisible by $p$ (and thus by $p^{2}$ ); this means that the multipliers $f(n)+k$ and $f(n)+\ell$ are also divisible by $p$, therefore $p \mid(f(n)+k)-(f(n)+\ell)=k-\ell$ as well.

On the other hand, if $f(k)-f(\ell)$ is divisible by $p$ but not by $p^{2}$, then choose the same number $D$ and set $n=p^{3} D-f(k)$. Then the positive numbers $f(k)+n=p^{3} D$ and $f(\ell)+n= p^{3} D+(f(\ell)-f(k))$ are respectively divisible by $p^{3}$ (but not by $p^{4}$ ) and by $p$ (but not by $p^{2}$ ). Hence in analogous way we obtain that the numbers $f(n)+k$ and $f(n)+\ell$ are divisible by $p$, therefore $p \mid(f(n)+k)-(f(n)+\ell)=k-\ell$.

We turn to the problem. First, suppose that $f(k)=f(\ell)$ for some $k, \ell \in \mathbb{N}$. Then by Lemma we have that $k-\ell$ is divisible by every prime number, so $k-\ell=0$, or $k=\ell$. Therefore, the function $f$ is injective.

Next, consider the numbers $f(k)$ and $f(k+1)$. Since the number $(k+1)-k=1$ has no prime divisors, by Lemma the same holds for $f(k+1)-f(k)$; thus $|f(k+1)-f(k)|=1$.

Now, let $f(2)-f(1)=q,|q|=1$. Then we prove by induction that $f(n)=f(1)+q(n-1)$. The base for $n=1,2$ holds by the definition of $q$. For the step, if $n>1$ we have $f(n+1)= f(n) \pm q=f(1)+q(n-1) \pm q$. Since $f(n) \neq f(n-2)=f(1)+q(n-2)$, we get $f(n)=f(1)+q n$, as desired.

Finally, we have $f(n)=f(1)+q(n-1)$. Then $q$ cannot be -1 since otherwise for $n \geq f(1)+1$ we have $f(n) \leq 0$ which is impossible. Hence $q=1$ and $f(n)=(f(1)-1)+n$ for each $n \in \mathbb{N}$, and $f(1)-1 \geq 0$, as desired.