$f,g \in \mathbb{Z}[x,y]$ satisfying: $\operatorname{Jac}(f,g)=0$ and $f,g \notin \mathbb{Z}[h]$ for every $h \in \mathbb{Z}[x,y]$? 
Is it possible to find $f,g \in \mathbb{Z}[x,y]$ (with $\deg(f),\deg(g) \geq 1$) such that the following two conditions are satisfied:
(1) $\operatorname{Jac}(f,g)=f_xg_y-f_yg_x = 0$.
(2) There exist no $h \in \mathbb{Z}[x,y]$ such that $f,g \in \mathbb{Z}[h]$.

Please see the answer to this question, in which it is shown that,
if in the above question we replace $\mathbb{Z}$ by some non-normal integral domain, then the answer is positive.
Any comments are welcome!
 A: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$No. As explained in this question, in $\mathbb{Q}[x,y]$, the condition $\operatorname{Jac}(f,g)=0$ implies that there exists an $h \in \mathbb{Q}[x,y]$ such that $f$ and $g$ are in $\mathbb{Q}[h]$. We now need some lemmas that are basically variants of Gauss's lemma, with multiplication replaced by composition.
Recall that a polynomial with coefficients in $\ZZ$ is called primitive if the set of its coefficients have $GCD=1$. I'll also define a polynomial to be very primitive if the GCD of the coefficients other than the constant term is $1$.
Lemma 1 Let $a \in \ZZ[t]$ be primitive and $b \in \ZZ[x,y]$ be very primitive. Then $a \circ b$ is primitive.
Proof: Suppose to the contrary that $p$ is a prime dividing every coefficient $a \circ b$. Let $\bar{a}$ and $\bar{b}$ denote the reductions modulo $p$, so these are polynomials in $(\ZZ/p)[t]$ and $(\ZZ/p)[x,y]$ respectively. The polynomial $\bar{a}$ is nonzero, $\bar{b}$ is not a constant, and $\ZZ/p$ is a field, so $\bar{a} \circ \bar{b}$ is nonzero. But the hypothesis is that $a \circ b$ is $0$ modulo $p$, and composition commutes with reduction modulo $p$. $\square$
Lemma 2: Let $b \in \ZZ[x,y]$ be very primitive, let $c \in \QQ[t]$ and suppose that $c \circ b \in \ZZ[x,y]$. Then $c \in \ZZ[t]$.
Proof: Write $c(t) = \tfrac{p}{q} a(t)$ with $a$ primitive and $p$ and $q \in \ZZ$ relatively prime. Then $c(b(x,y,)) = \tfrac{p}{q} a(b(x,y))$ and, by Lemma 1, $a(b(x,y))$ is primitive. So $q$ divides every coefficient of a primitive polynomial, and we deduce that $q=1$. So $c(t) = p a(t) \in \ZZ[t]$. $\square$
We now prove your result. Let $Jac(f,g)=0$. So there is $h \in \QQ[x,y]$ and $a$ and $b \in \QQ[t]$ such that $f=a \circ h$ and $g = b \circ h$. The case where $h$ is constant is clear, so we assume it is not. 
Subtracting a constant from $h$, (and changing $a$ and $b$ appropriately) we may assume that the constant term of $h$ is $0$. Rescaling $h$ by an appropriate element of $\QQ$ (and rescaling the coefficients of $a$ and $b$ correspondingly), we may assume that $h$ is primitive. A primitive polynomial with constant term $0$ is very primitive. So Lemma 2 tells us that $a$ and $b \in \ZZ[t]$, and we are done. $\square$
This argument generalizes immediately to any UFD, and with a bit more work to any normal ring.
