Is $x^j+x^k+2$ irreducible whenever $j+k$ is odd? Let $j,k$ be positive integers with $j>k$ and consider the polynomial 
$$f(x)=x^j+x^k+2$$
I want to prove the conjecture :

$f(x)$ is irreducible in $\mathbb Q[x]$, whenever $j+k$ is odd. This is true for $j\le 300$ as I checked with PARI/GP.

If $f$ has real roots, they obviously must be negative and the absolute value of any root must be less than $2$ for $j>2$. 
Moreover, $-1$ cannot be a root because of $f(-1)=2$, so $f(x)$ never can have a linear factor. 
Can we use the bound of the absolute values of the roots and that the constant coefficient is prime to show that $f(x)$ must be irreducible in $\mathbb Q[x]$

Tiniest bit of progress settling the case $j=k+1$.
By the usual business with Gauss' Lemma It suffices to show that there are no polynomials of positive degree $p(x),q(x)\in\Bbb{Z}[x]$ such that 
$$p(x)q(x)=f(x)\tag{1}$$
Because $f(0)=2$ we can, without loss of generality, assume that $p(0)=\pm1$ and $q(0)=\pm2$. Reducing $(1)$ modulo two gives us that
$$
\overline{p}(x)\overline{q}(x)=x^j+x^k=x^k(x^{j-k}+1).
$$
Our assumptions about the constant terms then allow us to deduce that
$$
\begin{aligned}
\overline{p}(x)&\mid x^{j-k}+1,\ \text{and}\\
x^k&\mid\overline{q}(x).
\end{aligned}
$$
In particular, it follows that $\deg p(x)\le j-k$.
The case $j=k+1$ can then be handled easily. We have seen that $p(x)$ is linear, so $f(x)$ has an odd integer root. The only alternatives are $x=\pm1$, and these were already excluded.
