It's not the roots, it's the "$2$"!
A polynomial is irreducible over a ring if it cannot be written as a product of two non-invertible polynomials. In $\mathbb{Z}$, "$2$" is noninvertible, so $(x^2+2)2$ is an appropriately "nontrivial" factorization.
Meanwhile, over in $\mathbb{Q}$, the polynomial "$2$" is invertible, since ${1\over 2}$ is rational (proof: exercise :P). So the factoriztion $(x^2+2)2$ is "trivial" in the context of $\mathbb{Q}$, since we can always extract a factor of $2$ from any polynomial.
EDIT: Think of it this way: saying that a polynomial is irreducible over a ring means it has no "nontrivial" factorizations. Now, when we make the ring bigger (e.g. pass from $\mathbb{Z}$ to $\mathbb{Q}$) two things happen:
So even though your first instinct might be "polynomials will only go from "irreducible" to "reducible" as the ring gets bigger," actually the opposite can happen!
In fact, here's a good exercise:
Can you find a polynomial $p\in\mathbb{Z}[x]$ which is irreducible over $\mathbb{Z}$ but reducible over $\mathbb{Q}$?
Note that the definition of reducibility over a field may sound different:
For $F$ a field, a polynomial $p\in F[x]$ is irreducible if $p$ cannot be written as the product of two nonconstant polynomials.
But this is actually equivalent to the definition I gave above, in case we're over a field: the noninvertible elements of $F[x]$ are precisely the nonconstant polynomials!
The statement
If $f(x)\in\mathbb{Z}[x]$ is reducible as element of $\mathbb{Q}[x]$ then it is reducible as element of $\mathbb{Z}[x]$
is indeed correct, but the converse is not generally true, because the polynomial $2x\in\mathbb{Z}[x]$ is reducible in $\mathbb{Z}[x]$, but it becomes irreducible in $\mathbb{Q}[x]$.
However, any nonzero polynomial over $\mathbb{Z}[x]$ can be written in the form
$$
cf(x)
$$
where $c\in\mathbb{Z}$, $f(x)=a_0+a_1x+\dots+a_nx^n\in\mathbb{Z}[x]$ and $\gcd(a_0,a_1,\dots,a_n)=1$: just compute the gcd of the coefficients and collect it. A polynomial such that the gcd of its coefficients is $1$ is called primitive.
For primitive polynomials, the converse implication also holds:
If $f(x)\in\mathbb{Z}[x]$ is primitive and it is reducible as an element of $\mathbb{Z}[x]$, then it is also reducible as an element of $\mathbb{Q}[x]$.
Let's see why. Suppose $f(x)=g(x)h(x)$, with both factors noninvertible in $\mathbb{Z}[x]$, but $f(x)$ is irreducible as an element of $\mathbb{Q}[x]$. Then one of the factors must be invertible in $\mathbb{Q}[x]$, so a constant polynomial and different from $\pm1$, because otherwise it would be invertible in $\mathbb{Z}[x]$. This would make $f(x)$ into a non primitive polynomial.
Usually the statement is written in terms of irreducibility:
A primitive polynomial $f(x)\in\mathbb{Z}[x]$ is irreducible as element of $\mathbb{Z}[x]$ if and only if it is irreducible as an element of $\mathbb{Q}[x]$.
The statement all primitive polynomials are irreducible is false. Consider $f(x)=x^2$: it is primitive, but reducible.
Best Answer
The element $2$ is a unit in $\mathbb Q[x]$ (it has inverse $\frac 12$) but not in $\mathbb Z[x]$ (since $\frac 12\notin \mathbb Z[x]$). We can write $$f(x) = 2(x^2+2).$$Since a polynomial $f$ is reducible iff it can be written as the product of two non-units, this means that $f$ is reducible in $\mathbb Z[x]$. However, this factorisation does not show that $f$ is reducible in $\mathbb Q[x]$ since $2$ is a unit. One can show in other ways (e.g. because $f$ has no rational roots) that $f$ is irreducible over $\mathbb Q$.
This condition is not arbitrary: it means that the ideal $(2x^2+4)$ is prime in $\mathbb Q[x]$ (and is equal to the maximal ideal ($x^2+2)$), but it is not prime in $\mathbb Z[x]$, since $2x^2+4\in(2x^2+4)$, but $$2,x^2+2\notin (2x^2+4).$$