$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\F{\mathbf{F}}$Let me provide a purely group-theoretic answer.

Let $H$ be a subgroup of finite index in $\mathrm{PSL}_2(\mathbf{Z})\simeq C_2\ast C_3$. Then, by Kurosh, $H$ is isomorphic to $C_2^{\ast e_2}\ast C_3^{\ast e_3}\ast \mathbf{Z}^{\ast e_0}$ for some integers $e_0,e_2,e_3\ge 0$ (clearly finite, and satisfying $e_2+e_3+e_0\ge 2$).

Since these are virtually free and since in a free group the index determines the rank, the triple $(e_2,e_3,e_0)$ determines the index $f(e_2,e_3,e_0)$ of $H$ in $\mathrm{PSL}_2(\mathbf{Z})$. Let me explicitly compute $f$.

For any groups $A,B$, we have $A\ast B\simeq A\ltimes B^{\ast A}$ (a kind of non-commutative wreath product). Hence, if $A'$ has index 2 in $A$, then $A\ast B$ has a subgroup of index 2 isomorphic to $A'\ltimes B^\ast {A'\sqcup A'}\simeq A'\ltimes (B\ast B)^{A'}\simeq A'\ast B\ast B$.

Given that $C_2^{e_2}$ has a subgroup of index 2 isomorphic to $\mathbf{Z}^{\ast e_2-1}$ for $e_2\ge 1$ (the kernel of the map onto $C_2$ that's identity on each factor), we deduce that $2f(e_2,e_3,e_0)=f(0,2e_3,2e_0+e_2-1)$ for $e_2\ge 1$.

Similarly, given that $C_3^{e_3}$, for $e_3\ge 1$ has a subgroup of index 3 isomorphic to $\mathbf{Z}^{\ast 2(e_3-1)}$, we deduce similarly that $3f(e_2,e_3,e_0)=f(3e_2,0,3e_0+2(e_3-1))$ whenever $e_3\ge 1$.

Hence $$f(e_2,e_3,e_0)=f(0,2e_3,2e_0+e_2-1)/2=f(0,0,6e_0+3(e_2-1)+2(2e_3-1))/6$$ $$=\frac16f(0,0,6e_0+3e_2+4e_3-5)$$
whenever $e_2,e_3\ge 1$; when one or both equals zero this is still valid, by direct verification and using that $f(0,0,1+k)=f(0,0,1+sk)/s$ by the values of ranks of subgroups of finite index in free groups.

Next, $\mathrm{PSL}_2(\mathbf{Z})$ has index 1 in itself, so $f(1,1,0)=1$. We just checked that $f(1,1,0)=f(0,0,2)/6$ by constructing a copy of index 6 of $F_2$. So $f(0,0,2)=6$, and, in turn, $f(0,0,n)=f(0,0,1+(n-1))=(n-1)f(0,0,2)=6(n-1)$. To conclude,

$$f(e_2,e_3,e_0)=\frac16f(0,0,6e_0+3e_2+4e_3-5)= 3e_2+4e_3+6(e_0-1).$$

When the generating rank is two, i.e., $e_0+e_2+e_3=2$ (excluding $e_2=2$ as $C_2^{\ast 2}$ can't occur), we get:
$$f(1,1,0)=1,\;f(1,0,1)=3;\;f(0,1,1)=4;f(0,2,0)=2;f(0,0,2)=6.$$

Hence subgroups of finite index of generating rank 2 in $\mathrm{PSL}_2(\mathbf{Z})$ are precisely those subgroups of index $1,2,3,4,6$. In index 1,2 there is a single such subgroup.

(Sequel: this was unfinished)

So it remains to describe subgroups of index 1,2,3,4,6 in $\PSL_2(\mathbf{Z})$.

Index 1 ($\simeq C_2\ast C_3$): the whole group (= principal congruence modulo 1).

Index 2 ($\simeq C_3\ast C_3$): there is a single one. It is congruence modulo 2, since $\PSL_2(\F_2)$ is a dihedral group of order 6.

Index 3 ($\simeq C_2\ast\mathbf{Z}$ or $\simeq C_2^{\ast 3}$): up to conjugacy there are two such subgroups. One is normal, actually $\simeq C_2^{\ast 3}$: it is congruence modulo 3, since $\PSL_2(\F_3)\simeq\mathrm{Alt}_4$ has a normal subgroup of index 3. The other ones are non-normal and congruence modulo 2, since $\PSL_2(\F_2)$ has a non-normal subgroup of index 3. [One representative consists of those matrices that are $\pm I_2$ modulo 3. If I'm correct it has a single conjugacy class of elements of order 2, so this group is isomorphic to $C_2\ast\mathbf{Z}$.]

Index 4 ($\simeq C_3\ast\mathbf{Z}$ or $\simeq C_2^{\ast 2}\ast C_3$): up to conjugacy there are two such subgroups, if I'm correct. One is congruence modulo 3, since $\PSL_2(\F_3)\simeq\mathrm{Alt}_4$ has a subgroup of index 4 (of order 3). [This one has no element of order 2, hence is isomorphic to $C_3\ast\mathbf{Z}$.] One is congruence modulo $4$, since $\PSL_2(\mathbf{Z}/4\mathbf{Z})$, which has order 48, has a subgroup of index 4 (of order 12), if I checked correctly.

Index 6 ($\simeq\mathbf{Z}^{\ast 2}$ or $\simeq C_3^{\ast 3}$ or $\simeq C_2^{\ast 4}$) is the most interesting case. I'll develop this later as it requires a number of verifications.