Is $x^j+x^k+2$ irreducible whenever $j+k$ is odd?

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 $\mathbb Q[x]$

Yes, that is indeed exactly how we can prove the irreducibility of $x^j+x^k+2$, by more closely inspecting its (complex!) roots.

Firstly, no root $\alpha \in \mathbb{C}$ can satisfy $|\alpha|<1$, since then we would have $$2=|\alpha^j+\alpha^k|\leq|\alpha|^j+|\alpha|^k<1+1=2.$$ We also cannot have $|\alpha|=1$ with odd $j+k$, since then $2=|\alpha^k||\alpha^{j-k}+1|=|\alpha^{j-k}+1|$ implies $\alpha^{j-k}=1$ (for example by looking at the complex plane). However from $(\alpha^{j-k}+1)\alpha^k=-2$ we have $\alpha^k=-1$, a contradiction with $\alpha$ being an odd root of unity.

So we have $|\alpha|>1$, and the irreducibility now follows in a standard way: assuming a monic factor $p(x)$ with constant coefficient equal $\pm 1$ (we can choose such since $2$ is a prime), product of its roots $\alpha_i$ (which are also roots of the original polynomial) are given by $1=|p(0)|=\prod |\alpha_i|>1$, a contradiction.

Remark: If $j+k$ was even, then there would indeed be roots on the unit circle and we could not use this reasoning. You have encountered one such root, namely $\alpha=-1$. Another one would be for example $\alpha=i$ in $x^6+x^2+2=(x^2+1)(x^4-x^2+2).$