By considering this equation modulo $3$, we get
$$5^{a} + 3^{b} - 2^{c} \equiv 2^{a} - 2^{c} \equiv 2 \pmod{3}$$
This occurs only when $a$ is even and $c$ is odd.

If $c = 1$, the equation becomes $5^{a} + 3^{b} = 34$, which is equivalent to $a = 2$ and $b = 1$.

If $c \neq 1$, then $c \geq 3$, since $c$ is odd. By considering this equation modulo $8$, and using the fact that $a$ is even, we get
$$1 + 3^{b} \equiv 0 \pmod{8}$$
which can never occur since $3^{b}$ is congruent to either $1$ or $3$ modulo $8$.

Hence, the unique solution to this equation is $(a, b, c) = (2, 1, 1)$.