First we perform the following substitutions on the original relation:

1. With $a = b = 1$, we find that $f(1) + 1 \mid f(1)^{2} + 1$, which implies $f(1) = 1$.

2. With $a = 1$, we find that $b + 1 \mid f(b) + 1$. In particular, $b \leq f(b)$ for all $b \in \mathbb{Z}^{+}$.

3. With $b = 1$, we find that $f(a) + 1 \mid a^{2} + f(a)$, and thus $f(a) + 1 \mid a^{2} - 1$. In particular, $f(a) \leq a^{2} - 2$ for all $a \geq 2$.

Now, let $p$ be any odd prime. Substituting $a = p$ and $b = f(p)$ in the original relation, we find that $2 f(p) \mid p^{2} + f(p) f(f(p))$. Therefore, $f(p) \mid p^{2}$. Hence the possible values of $f(p)$ are $1$, $p$ and $p^{2}$. By (2) above, $f(p) \geq p$ and by (3) above $f(p) \leq p^{2} - 2$. So $f(p) = p$ for all primes $p$.

Substituting $a = p$ into the original relation, we find that $b + p \mid p^{2} + p f(b)$. However, since $(b + p)(f(b) + p - b) = p^{2} - b^{2} + b f(b) + p f(b)$, we have $b + p \mid b f(b) - b^{2}$. Thus, for any fixed $b$ this holds for arbitrarily large primes $p$ and therefore we must have $b f(b) - b^{2} = 0$, or $f(b) = b$, as desired.

As above, we have relations (1)-(3). In (2) and (3), for $b = 2$ we have $3 \mid f(2) + 1$ and $f(2) + 1 \mid 3$. These imply $f(2) = 2$.

Now, using $a = 2$ we get $2 + b \mid 4 + 2 f(b)$. Let $f(b) = x$. We have
$$\begin{array}{r} 1 + x \equiv 0 \quad (\bmod b + 1) \\ 4 + 2x \equiv 0 \quad (\bmod b + 2) \end{array}$$
From the first equation $x \equiv b (\bmod b + 1)$ so $x = b + (b + 1)t$ for some integer $t \geq 0$. Then
$$0 \equiv 4 + 2x \equiv 4 + 2(b + (b + 1)t) \equiv 4 + 2(-2 - t) \equiv -2t \quad (\bmod b + 2)$$
Also $t \leq b - 2$ because $1 + x \mid b^{2} - 1$ by (3).

If $b + 2$ is odd, then $t \equiv 0 (\bmod b + 2)$. Then $t = 0$, which implies $f(b) = b$.
If $b + 2$ is even, then $t \equiv 0 (\bmod (b + 2)/2)$. Then $t = 0$ or $t = (b + 2)/2$. But if $t \neq 0$, then by definition $(b + 4)/2 = (1 + t) = (x + 1)/(b + 1)$ and since $x + 1 \mid b^{2} - 1$, then $(b + 4)/2$ divides $b - 1$. Therefore $b + 4 \mid 10$ and the only possibility is $b = 6$. So for even $b$, $b \neq 6$ we have $f(b) = b$.

Finally, by (2) and (3), for $b = 6$ we have $7 \mid f(6) + 1$ and $f(6) + 1 \mid 35$. This means $f(6) = 6$ or $f(6) = 34$. The latter is discarded as, for $a = 5$, $b = 6$, we have by the original equation that $11 \mid 5(5 + f(6))$. Therefore $f(n) = n$ for every positive integer $n$.

We proceed by induction. As in Solution 1, we have $f(1) = 1$. Suppose that $f(n - 1) = n - 1$ for some integer $n \geq 2$.

With the substitution $a = n$ and $b = n - 1$ in the original relation we obtain that $f(n) + n - 1 \mid n^{2} + f(n)(n - 1)$. Since $f(n) + n - 1 \mid (n - 1)(f(n) + n - 1)$, then $f(n) + n - 1 \mid 2n - 1$.

With the substitution $a = n - 1$ and $b = n$ in the original relation we obtain that $2n - 1 \mid (n - 1)^{2} + (n - 1)f(n) = (n - 1)(n - 1 + f(n))$. Since $(2n - 1, n - 1) = 1$, we deduce that $2n - 1 \mid f(n) + n - 1$.

Therefore, $f(n) + n - 1 = 2n - 1$, which implies the desired $f(n) = n$.