Let $3^a + 3^b + 3^c = m^2$. Since $m$ is odd we have $m^2 \equiv 1 \pmod{8}$. For each positive integer $k$ we have $3^k \equiv 1$ or $3^k \equiv 3 \pmod{8}$. Therefore, $3^a \equiv 3^b \equiv 3^c \equiv 1 \pmod{8}$ and we get that $a$, $b$ and $c$ are odd numbers. Let $a \leq b \leq c$. Then $3^a (3^{b-a} + 3^{c-a} + 1) = m^2$. Now since $a$ is odd, we get that $3^{b-a} + 3^{c-a} + 1 \equiv 0 \pmod{3}$. Thus, $a = b = c$ and we get all solutions: $(a, b, c) = (2k-1, 2k-1, 2k-1)$, where $k$ is a positive integer.