Showing $x^4 + 1$ is irreducible in $\mathbb{Q}[x]$. Clearly, none of the roots are in $\mathbb{Q}$ so $f(x) = x^4 + 1$ does not have any linear factors. Thus, the only thing left to check is to show that $f(x)$ cannot reduce to two quadratic factors.
My proposed solution was to state that $f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$ but $\pm i \not\in \mathbb{Q}$ so $f(x)$ is irreducible.
However, I stumbled across this post $x^4 + 1$ reducible over $\mathbb{R}$... is this possible? with a comment suggesting that $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$ which turns out to be a case that I did not fully consider. It made me realize that $\mathbb{Q}[x]$ being a UFD only guarantees a unique factorization of irreducible elements in $\mathbb{Q}[x]$ (which $x^2 \pm i$ nor $x^2 \pm \sqrt{2} x + 1$ aren't in $\mathbb{Q}[x]$) so checking a single combination of quadratic products is not sufficient.
Therefore, what is the ideal method for checking that $x^4 + 1$ cannot be reduced to a product of two quadratic polynomials in $\mathbb{Q}[x]$? Am I forced to just brute force check solutions of $x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$ don't have rational solutions $(a,b,c,d) \in \mathbb{Q}^4$?

You can indeed try a factorization of the form
$x^4+1=(x^2+ax+b)(x^2+cx+d).$
Expanding the right side and matching terms with like powers gives
$x^3$ terms: $a+c=0,c=-a$
$x^2$ terms: $ac+b+d=0,b+d=a^2$
$x^1$ terms: $ad+bc=a(d-b)=0,a=0$ or $d=b$
The case $a=0$ leads to $(x^2+i)(x^2-i)$ which fails to lie in $\mathbb Q[x]$. So we try $d=b$, which then means $d=b=a^2/2$ from the matching of $x^2$ terms. Then matching the $x^0$ terms gives:
$x^0$ terms: $bd=b^2=1.$
Then $b=a^2/2=\pm1$ and neither choice of the $\pm$ sign allows a rational value for $a$. In fact only $b=+1,a=\pm\sqrt2$ admits a quadratic-quadratic factorization even over the reals.
