$f(x) = x + 1$ for all $x\in \mathbb{Z}$.

Define $S$ as the set of all integers $x$ such that $f(x) = x + 1$. First, observe that if $x\in S$, then $f(x + 1) = f(f(x)) = x(x + 1) - x^2 + 2 = x + 2$, so $x + 1\in S$. Since $S\subseteq \mathbb{Z}$, then either $S = \mathbb{Z}$, $S$ is empty, or $S = \{x \in \mathbb{Z} | x \geq M\}$ for some integer $M$. However, by (1) we can see that $f(f(1)) = f(1) + 1$, so $S$ is non-empty. Suppose for the sake of contradiction that $S = \{x \in \mathbb{Z} | x \geq M\}$ for some integer $M$. Obviously, $M \leq f(1)$. By definition of $S$, we have $f(x) = x + 1$ if and only if $x \geq M$. We will proceed with lemmas. Lemma 1For all integers $x$, if $x < M$ and $x \neq 1$, then $f(x) < M$. ProofIf $f(x) \geq M$, by (1) we get: \begin{align*} f(x) + 1 &= xf(x) - x^2 + 2 \\ (x - 1)(f(x) - x - 1) &= 0 \end{align*}so either $x = 1$ or $x\in S$ i.e. $x \geq M$. Lemma 2$M \geq 3$. ProofSuppose for the sake of contradiction that $M \leq 2$. By $(1)$, $f(f(0)) = 2$, so $$ 3 = f(f(f(0))) = f(0)f(f(0)) - f(0)^2 + 2 = 2f(0) - f(0)^2 + 2 = 3 - (f(0) - 1)^2 $$As a result, $f(0) = 1$, so $M \leq 0$. Now, we have $f(M - 1) < M$ by Lemma 1 as $M - 1 < M < 1$. But this also means $f(f(M - 1)) < M$ by Lemma 1, while we have $$ f(f(M - 1)) = (M - 1)(f(M - 1) - (M - 1)) + 2 \geq 2 > M + 1 $$since $f(M - 1) \leq M - 1 < 0$. Contradiction. Lemma 3$f(x) \neq 1$ for all $x\in \mathbb{Z}$. ProofOtherwise, if $f(t) = 1$ for some $t\in \mathbb{Z}$, (1) yields $f(1) = t - t^2 + 2 \leq 2 < M$; a contradiction as $f(1) \geq M$. Finally, consider a sequence of integers $(a_k)_{k \geq 0}$ where $a_0 = -1$ and $a_{k+1} = f(a_k)$ for each $k \geq 0$. By the property of such sequence, it can be easily proved that either all terms are distinct in value or the sequence will eventually be periodic. Furthermore, Lemma 3 implies that no term of the sequence is equal to $1$, and thus by Lemma 1, all of them are less than $M$; i.e. $a_k < M$ for all $k \geq 0$. By (1), we have that for all $k\geq 0$, \begin{align*} a_{k + 2} &= a_k a_{k + 1} - a_k^2 + 2 \\ &= a_k (a_{k+1} - a_k) + 2 \\ a_{k + 2} - a_{k + 1} - 1 &= a_k a_{k + 1} - a_k^2 - a_{k + 1} - 1 \\ &= (a_k - 1)(a_{k + 1} - a_k - 1) \end{align*} First, we prove that $(a_k)_{k \geq 0}$ contains infinitely many non-negative terms. Suppose for the sake of contradiction that it is not. Let $K$ be a non-negative integer such that $a_k < 0$ for all $k \geq K$. Then, for each $k \geq K$, we have $$ a_k (a_{k + 1} - a_k) + 2 = a_{k + 2} < 0 $$and since $a_k < 0$, then $a_{k + 1} > a_k$. But then this means $a_K, a_{K + 1}, \ldots$ is a strictly increasing sequence of negative integers. Contradiction. Hence, $(a_k)_{k \geq 0}$ indeed contains infinitely many non-negative terms. But the sequence is bounded above, so there exists indices $i < j$ with $a_i = a_j$. This implies that $(a_k)_{k \geq 0}$ is eventually periodic (as $a_{k+1} = f(a_k)$ for each $k \geq 0$). Now, letting $b_k = a_{k + 1} - a_k - 1$, since $(a_k)_{k \geq 0}$ is eventually periodic, so is $(b_k)_{k \geq 0}$ and $(|b_k|)_{k \geq 0}$ as well. In particular, it is bounded. But the above equation tells us that $b_{k + 1} = (a_k - 1) b_k$ for all $k \geq 0$, and $a_k \neq 1$ for all $k\geq 0$ implies that the sequence $(|b_k|)_{k \geq 0}$ is not only bounded, but is non-decreasing. Such sequence, consisting only of integers, must be eventually constant. Furthermore, since $a_0 = -1 < M$, we have $a_1 \neq 0$ so $b_0 \neq 0$. That means the constant is non-zero. Then, for all $k$ big enough, we have $|b_{k+1}| = |b_k|$, $|a_k - 1| = 1$, and thus $a_k \in \{0, 2\}$. From this, we show that $f(2) = 2$ and in fact $a_k = 2$ for every $k$ big enough. Consider a positive integer $k'$ big enough such that $a_k \in \{0, 2\}$ for all $k \geq k'$. Suppose for the sake of contradiction that $a_{k + 2} = 0$ for some $k \geq k'$. Since $f(f(0)) = 2$, this means $a_k = 2$, so $f(f(2)) = 0$. But then (1) yields $f(2) = 1$; a contradiction to Lemma 3. Hence, $a_k = 2$ for all $k \geq k' + 2$, and thus $f(2) = 2$. But then, $b_k = a_{k + 1} - a_k - 1 = -1$ for all $k$ large enough. Since $(|b_k|)_{k \geq 0}$ is a non-decreasing sequence of positive integers, it must be constant; i.e. $|b_0| = |b_1| = 1$. But $b_1 = (a_0 - 1)b_0 = -2b_0$; a contradiction. Hence, $S = \mathbb{Z}$ and $f(x) = x + 1$ for all $x\in \mathbb{Z}$.