No, there is no such polynomial. Any polynomial $f(x) \in \mathbb{Z}[x]$ with degree greater than $1$ is reducible modulo every prime factor of every value it takes.
For, take any value of $n$ for which $f(n) \neq \pm 1$. (There must exist such $n$ because $f$ can take the values $1$ and $-1$ only finitely many times.) Consider any prime factor $p$ of $f(n)$. Then $f(n) \equiv 0 \mod p$, which means that $f$ is reducible in $\mathbb{F}_p[x]$: it is divisible by the polynomial $x-n$.
There are two crucial results here.
Dedekind's theorem: Let $f$ be a monic irreducible polynomial over $\mathbb{Z}$ of degree $n$ and let $p$ be a prime such that $f$ has distinct roots $\bmod p$ (this is true for precisely the primes not dividing the discriminant). Suppose that the prime factorization of $f \bmod p$ is
$$f \equiv \prod_{i=1}^k f_i \bmod p.$$
Then the Galois group $G$ of $f$ contains an element of cycle type $(\deg f_1, \deg f_2, ...)$. In particular, if $f$ is irreducible $\bmod p$, then $G$ contains an $n$-cycle.
Frobenius density theorem: The density of the primes with respect to which the factorization of $f \bmod p$ has the above form is equal to the density of elements of $G$ with the corresponding cycle type. In particular, for every cycle type there is at least one such prime $p$.
It follows that
$f$ is reducible $\bmod p$ for all $p$ if and only if $G$ does not contain an $n$-cycle.
The smallest value of $n$ for which this is possible is $n = 4$, where the Galois group $V_4 \cong C_2 \times C_2$ has no $4$-cycle. Thus to write down a family of examples it suffices to write down a family of irreducible quartics with Galois group $V_4$. As discussed for example in this math.SE question, if
$$f(x) = (x^2 - a)^2 - b$$
is irreducible and $a^2 - b$ is a square, then $f$ has Galois group $V_4$. In particular, taking $b = a^2 - 1$ the problem reduces to finding infinitely many $a$ such that
$$f(x) = x^4 - 2ax^2 + 1$$
is irreducible. We get your examples by setting $a = 0, 5$.
By the rational root theorem, the only possible rational roots of $f$ are $\pm 1$, so by taking $a \neq 1$ we already guarantee that $f$ has no rational roots. If $f$ splits into two quadratic factors, then they both have constant term $\pm 1$, so we can write
$$x^4 - 2ax + 1 = (x^2 - bx \pm 1)(x^2 + bx \pm 1)$$
for some $b$. This gives
$$2a = b^2 \mp 2.$$
Thus $f$ is irreducible if and only if $2a$ cannot be written in the above form (and also $a \neq 1$).
Classifying polynomials $f$ with this property seems quite difficult in general. When $n = 4$, it turns out that $V_4$ is in fact the only transitive subgroup of $S_4$ not containing a $4$-cycle, but for higher values of $n$ there should be lots more, and then one has to tell whether a polynomial has one of these as a Galois group...
(Except if $n = q$ is prime; in this case $q | |G|$ so it must have a $q$-cycle.)
Best Answer
The field $\mathbf{F}_{q^2}$ contains the square roots of all of the elements of $\mathbf{F}_q$, so the splitting field of $f$ over $\mathbf{F}_q$ is $\mathbf{F}_{q^2}$, a degree 2 extension.
Thus, the factorization of $f$ consists only of linear and quadratic factors -- in particular, no quartic factors -- thus it is not irreducible over $\mathbf{F}_q$.