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.
I intended to post the following as a comment but it is too long:
A very easy argument showing that every prime $p>n+1$ works for the injectivity of the reduction modulo $p$ of a finite subgroup $H$ of $SL_n(\mathbb Z)$ is as follows: Since every element of $H$ is finite, its characteristic polynomial is a product of cyclotomic polynomials. Reductions of cyclotomic polynomials modulo $p$ with order (defined as the order of an underlying root of unity in the multiplicative group of invertible elements in $\mathbb C$) prime to $p$ are never congruent to a power of $(1-x)$. Cyclotomic polynomials of order divisible by $p$ are of degree at least $p-1$. Since a non-trivial element $h\in H$ is diagonalisable, its characteristic polynomial is not a power of $(1-x)$. The reduction of this characteristic polynomial modulo $p$ is thus also distinct from a power of $(1-x)$. This implies that $h$ modulo $p$ is non-trivial in $SL_n(\mathbb F_p)$.