It depends on how much group theory you want to use. If $G$ is such a simple group and $\chi$ is a faithful complex irreducible character of degree $3$, then a Theorem of Feit and Thompson proves that $|G|$ is not divisible by any prime $p > 7$. It is easy to check (since $Z(G)$ contains no element of order $3$), that $G$ has Abelian Sylow $3$-subgroups. Hence for any $x \in G$ of order $3$, we have ${\rm gcd}([G:C_{G}(x)],\chi(1)) = 1$. By a result of Burnside, we have $\chi(x) = 0$ since $G$ is simple. It then easily follows that $|G|$ is not divisible by $9$. It also follows that a Sylow $3$-subgroup of $G$ is self-centralizing ( from this, it already follows from 1962(?) theorem of Feit and Thompson in Nagoya J. Math, that $G \cong {\rm PSL}(2,7)$ or $A_{5}$. For,, more generally, we see that $\chi(g) = 0 $ whenever $g$ centralizes a Sylow $3$-subgroup $P$ of $G$, and then $|C_{G}(P)|$ divides $3$). A representation theoretic proof in this special case can be sketched as follows (using a few tricks not available to Blichfeldt):

Now $G$ has cyclic Sylow $7$-subgroups for otherwise, $G$ contains a element of order $7$ with eigenvalues $1,\omega, \omega^{-1}$, where $\omega = e^{\frac{2 \pi i}{7}}$, which contradicts a theorem of Blichfeldt ( as $\chi$ must be primitive ( otherwise $G$ would be solvable)). Consideration of reduction (mod $7$) shows that the Sylow $7$-subgroup of $G$ has order at most $7$. Furthermore, if $7$ divides $|G|$, it follows from Burnside's normal $p$-complement theorem that a Sylow $7$-normalizer must have order $21$.

Since $G$ has no element of order $35$, it follows from another theorem of Blichfeldt that $|G|$ can't be divisible by $35$ ( since otherwise, $\chi$ is neither $5$-rational nor $7$-rational, and would contain an element of order $35$. But consideration of reduction (mod $5$) shows that no element of order $5$ can commute with any element of order $7$).

If $G$ contains an element of order $5$ with only two eigenvalues then an argument of Blichfeldt shows that ${\rm SL}(2,5)$ as a subgroups, and contains an element of order $6$ with eigenvalues $1, \alpha, {\bar \alpha}$ for a primitive $6$-th root of unity, which ( by another of his results) contradicts the primitivity of $\chi$. It follows that $G$ has cyclic Sylow $5$-subgroups which have order at most $5$ on consideration of reduction (mod $5$).

It now follows that $|G|$ has the form $2^{a}.3.5$ or $2^{b}.3.7$ for integers $a,b$. Similarly to the argument for $7$, we may conclude that a Sylow $5$-normalizer has order $10$ if $5$ divides $|G|$. This gives $a \equiv 2$ (mod $4$) or $b \equiv 3$ (mod $6$) by Sylow's theorem. A Sylow $2$-subgroup $S$ of $G$ has an Abelian normal subgroup $A$ of index dividing $2$, and a 1965 Theorem of Brauer shows that $|A|$ divides $4$ so $|S \leq 8$ and we do get $|G| = 60$ or $168$.

It depends on how much group theory you want to use. If $G$ is such a simple group and $\chi$ is a faithful complex irreducible character of degree $3$, then a Theorem of Feit and Thompson proves that $|G|$ is not divisible by any prime $p > 7$. It is easy to check (since $Z(G)$ contains no element of order $3$), that $G$ has Abelian Sylow $3$-subgroups. Hence for any $x \in G$ of order $3$, we have ${\rm gcd}([G:C_{G}(x)],\chi(1)) = 1$. By a result of Burnside, we have $\chi(x) = 0$ since $G$ is simple. It then easily follows that $|G|$ is not divisible by $9$. It also follows that a Sylow $3$-subgroup of $G$ is self-centralizing ( from this, it already follows from 1962(?) theorem of Feit and Thompson in Nagoya J. Math, that $G \cong {\rm PSL}(2,7)$ or $A_{5}$. For, more generally, we see that $\chi(g) = 0 $ whenever $g$ centralizes a Sylow $3$-subgroup $P$ of $G$, and then $|C_{G}(P)|$ divides $3$). A representation theoretic proof in this special case can be sketched as follows (using a few tricks not available to Blichfeldt):

Now $G$ has cyclic Sylow $7$-subgroups for otherwise, $G$ contains a element of order $7$ with eigenvalues $1,\omega, \omega^{-1}$, where $\omega = e^{\frac{2 \pi i}{7}}$, which contradicts a theorem of Blichfeldt ( as $\chi$ must be primitive ( otherwise $G$ would be solvable)). Consideration of reduction (mod $7$) shows that the Sylow $7$-subgroup of $G$ has order at most $7$. Furthermore, if $7$ divides $|G|$, it follows from Burnside's normal $p$-complement theorem that a Sylow $7$-normalizer must have order $21$.

Since $G$ has no element of order $35$, it follows from another theorem of Blichfeldt that $|G|$ can't be divisible by $35$ ( since otherwise, $\chi$ is neither $5$-rational nor $7$-rational, and would contain an element of order $35$. But consideration of reduction (mod $5$) shows that no element of order $5$ can commute with any element of order $7$).

If $G$ contains an element of order $5$ with only two eigenvalues then an argument of Blichfeldt shows that ${\rm SL}(2,5)$ as a subgroups, and contains an element of order $6$ with eigenvalues $1, \alpha, {\bar \alpha}$ for a primitive $6$-th root of unity, which ( by another of his results) contradicts the primitivity of $\chi$. It follows that $G$ has cyclic Sylow $5$-subgroups which have order at most $5$ on consideration of reduction (mod $5$).

It now follows that $|G|$ has the form $2^{a}.3.5$ or $2^{b}.3.7$ for integers $a,b$. Similarly to the argument for $7$, we may conclude that a Sylow $5$-normalizer has order $10$ if $5$ divides $|G|$. This gives $a \equiv 2$ (mod $4$) or $b \equiv 3$ (mod $6$) by Sylow's theorem. A Sylow $2$-subgroup $S$ of $G$ has an Abelian normal subgroup $A$ of index dividing $2$, and a 1965 Theorem of Brauer shows that $|A|$ divides $4$ so $|S| \leq 8$ and we do get $|G| = 60$ or $168$.