The subgroups of ${\rm PSL}(2,q)$ for prime powers $q$ were classified by Dickson in 1901. You can find the lists and proofs in Huppert's German book "Endliche Gruppen I". Of course you need the subgroups of ${\rm PGL}(2,p)$, which is twice as big, and then you need to check their inverse images in ${\rm GL}(2,p)$ for irreducibility. So you have some work to do!
I will have a go at listing the minimal irreducible groups, but it is a little complicated, and I may not get it absolutely right.
For odd primes $r$ dividing $p+1$, the cyclic group of order $r$.
The cyclic group of order $2^r$, where $2^{r-1}$ is the highest power of 2 dividing $p-1$.
The dihedral group of order $2r$ for odd primes $r$ dividing $p-1$.
I believe that is all when $p \equiv 3 \pmod 4$. (That case is simpler, because all elements of order 4 act irreducibly.) It is more complicated when
$p \equiv 1 \pmod 4$. In that case, you also get:
The dihedral group $D_8$.
The quaternion group $Q_8$.
For odd primes $r$ dividing $p-1$, a group of order $4r$ contained in ${\rm SL}(2,p)$.
This is the complicated one. After doing some computations with $p=17$, I think, for all powers $2^r$ dividing $p-1$ with $r \ge 3$, there is a nonabelian group of order $2^{r+1}$ intersecting the scalar subgroup with order $2^{r-1}$.
Here is a bit more information on methods used. The maximal subgroups of ${\rm GL}(2,p)$ (with $p$ prime) are the imprimitive groups of order $2(p-1)^2$, the semilinear groups of order $2(p^2-1)$, groups whose intersection with ${\rm SL}(2,p)$ is a double cover of $A_4$, $S_4$ or $A_5$, and groups containing ${\rm SL}(2,p)$. Since ${\rm SL}(2,p)$ itself is never minimal irreducible, we can forget about that, and the ones involving $A_4$, $S_4$ or $A_5$ all have a normal irredcuible subgroup $Q_8$, which is itself imprimitive and/or semilinear, so we can effectively forget about those.
So we can restrict attention to imprimitive and semilinear subgroups. The maximal semilinears have a cyclic subgroup of order $q^2-1$ which acts irreducibly but not absolutely irreducibly, and it is not hard to see that their minimal irreducible subgroups are those in classes 1 and 2 above.
So it remains to find the minimal irreducible subgroups of the maximal imprimitive subgroups. These are all conjugate, and one of them, $H$ say, consists of the monomial matrices, so we can just look for subgroups of $H$. So the group $D$ of diagonal matrices has index 2 in $H$ and is reducible. So a minimal irreducible $K$ will have a subgroup of $D$ of index 2. If $|K \cap D|$ is odd, then get the groups in class 3 above. As I noted before, if $p \equiv 3 \pmod 4$, then elements of order 4 act irreducible, and any other such subgroup would have an element of order 4, so would not be minimal irreducible. But it is more complicated when $p \equiv 1 \pmod 4$.
