Adding my bit with a view of covering the possibility (somewhat implied by OP's comments) that the task really is to find all the irreducible quartic polynomials over $\Bbb{F}_2$ using the theorem that they must all be factors of $p(x)=x^{16}-x$.
The given minimal polynomial of $\alpha$, $f_1(x)=x^4+x^3+1$ is obviously one of those. A useful trick for finding another one is to go to the so called reciprocal polynomial, i.e.
$$
f_2(x)=x^4f_1(\frac1x)=x^4(\frac1{x^4}+\frac1{x^3}+1)=1+x+x^4.
$$
We immediately see that
$$
f_2(\frac1\alpha)=\alpha^4f_1(\alpha)=0,
$$
so $f_2(x)$ is a multiple of the minimal polynomial of $1/\alpha$. As the elements $\alpha$ and $1/\alpha$ generate the same extension field, their respective minimal polynomials must be of the same degree. Therefore $f_2(x)$ is
the minimal polynomial of $1/\alpha$ and is, thus also irreducible.
Let's take stock. We know that in addition to $f_1(x)$ and $f_2(x)$ the polynomial $p(x)$ is divisible by $q(x)=x^4-x$ (if you don't know how to deduce this from properties of finite fields, then you can just calculate that
$q(x)^4+q(x)=p(x)$. Thus we know that $p(x)$ factors like
$$
p(x)=q(x) f_1(x) f_2(x) r(x),
$$
where $r(x)$ is some yet unknown factor. We can partially factor $p(x)$ directly as follows:
$$
p(x)=x(x^{15}-1)=x(x^5-1)(x^{10}+x^5+1)=x(x-1)(x^4+x^3+x^2+x+1)(x^{10}+x^5+1),
$$
and $q(x)$ as follows:
$$
q(x)=x(x^3-1)=x(x-1)(x^2+x+1),
$$
where that last quadratic factor is irreducible.
I leave it as an exercise for you to check that
$$
(x^2+x+1)f_1(x)f_2(x)=x^{10}+x^5+1.
$$
Putting all this together gives us that the mystery factor $r(x)=f_3(x)=x^4+x^3+x^2+x+1$.
I next claim that $f_3(x)$ is irreducible. As it has no zeros in
$\Bbb{F}_2$ it has no linear factors in $\Bbb{F}_2[x]$. It cannot be a product of two distinct quadratics, because then it should be a factor of $q(x)$.
It cannot be the square of a quadratic, because then its zeros would not be simple, but $f_3(x)$ is a factor of $p(x)$ that has 16 distinct zeros.
Thus the list $f_1(x), f_2(x), f_3(x)$ is a complete list of irreducible quartics in $\Bbb{F}_2[x].$
Yet another way of deducing the irreducibility of $f_3(x)$ is to observe that
$f_3(x)\mid x^5-1$, so its zeros are fifth roots of unity. If you are familiar with the cyclicity of the multiplicative groups of finite fields, then you can immediately check that $\Bbb{F}_{16}$ has no proper subfields containing primitive fifth roots of unity. Therefore the minimal polynomial of a fifth root of unity over $\Bbb{F}_2$ must have degree four, i.e. equal to $f_3(x)$.
Observe that $f_3(x)$ is its own reciprocal polynomial. Polynomials with this property are called palindromic because you can equally well read their sequence of coefficients backwards. Therefore our first trick won't give us a fourth irreducible quartic.