$c = 0$ is clearly impossible. Notice that $1788^c \equiv 0 \mod{4} \implies a^{12} + 3^b \equiv 0 \mod{4}$. $a$ is clearly odd, so $a^{12} \equiv 1 \mod{4} \implies 3^b \equiv 3\mod{4}$, so $b$ is odd. Notice $a^{12} \equiv 1 \mod{8}$, $3^b \equiv 3\mod{8}$, so $1788^c \equiv 4 \mod{8} \implies c = 1$. So, $1788 = a^{12} + 3^b$, which clearly has no solutions. ($a = 1$ fails and $a > 2$ means $a^{12} > 1788$)

In Azerbaijan,note that $0\notin\mathbb{N}$ $c=1$ is trivial (Note:They cut off my 2 points because I wrote $c\neq1$ is obvious ) For $c\geq2$, $1788^c\equiv0\mod8$. If $3^b\equiv3\mod8$, then $(a^6)^2\equiv5\mod8$ which is not possible. If $3^b\equiv1\mod8$, then $(a^6)^2\equiv7\mod8$ which is not possible.