$X^4-4X^2-1$ irreducible over $\mathbb{Q}[X]$

Your polynomial has no rational roots. Therefore, if it was reducible in $\mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)\in\mathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.

The roots of your polynomial are $\pm\sqrt{\sqrt5+2}$ and $\pm i\sqrt{\sqrt5-2}$. Of these, only the last two are complex non-real. But$$\left(X-i\sqrt{\sqrt5-2}\,\right)\left(X+i\sqrt{\sqrt5-2}\,\right)=X^2-2+\sqrt5\notin\mathbb{Q}[X].$$Therefore, your polynomial is irreducible.

Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.

The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.

More generally, a similar argument shows that if $e(x)\in\mathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)\in \mathbb Z[x]$ , is $f(x^2)$ also irreducible in $\mathbb Z[x]$? for more details.