A slightly different argument showing that every finite subgroup of $SL_2(\mathbb Z)$ is of cardinality a divisor of $24$ goes as follows: Consider such a finite subgroup $H$. Since the coefficients of all elements of $H$ involve only a finite number of prime divisors, the obvious group homomorphism from $SL_2(\mathbb Z)$ into $SL_2(\mathbb F_p)$
where $\mathbb F_p$ is the finite field with cardinality a prime number $p$ is injective for 
almost all primes. Since  $SL_2(\mathbb F_p)$ has $p(p^2-1)$ elements, the cardinality $h$
of the finite group $H$ divides $p(p^2-1)$ for almost all prime numbers. This implies that 
$h$ divides $24$. Indeed, quadratic reciprocity shows that $2$ and $3$ are the only possible prime divisors of $h$ and gives upper bounds on the maximal exponents $\alpha,\beta$ such that 
$2^\alpha\cdot 3^\beta$ divides $p(p^2-1)$ almost all primes.