Is $x^4+2x^2-8m^2x+1$ irreducible over $\mathbb{Q}$?

Clearly there are no integer (hence no rational roots), so we need to only check for quadratic factors. So we need to only check for the form $$x^4+2x^2-8m^2x+1=(x^2+ax\pm1)(x^2-ax\pm1)$$ which gives $2-a^2=2$ and $\pm2a=-8m^2$ which is possible only when $a=m=0$.

Its irreducibility follows nicely from Perron's criterion when applied to reciprocal $g(x)=x^4f(1/x)=x^4-8m^2x^3+2x^2+1$:

Perron's criterion: Let $P(x) = x^n + a_{n-1} x^{n-1} + \dots + a_0 \in \mathbb{Z}[x]$. If $|a_{n-1}| > 1 + |a_{n-2}| + \dots + |a_0|$, then $P(x)$ is irreducible.

Since $8m^2> 1+2+1 = 4$ clearly holds for $m>0$, it follows $g(x)$ is irreducible, and particularly $f(x)$ is irreducible as well.