Rmc, 2020

We can assume $a \le b \le c$. If $a \ge 2$, then $2^a + 2^b + 2^c + 3 \equiv 3 \pmod{4}$, hence it cannot be a square. We deduce that $a = 1$, and we must find $b \le c$ such that $2^b + 2^c + 5$ is a square.

If $b \ge 3$, then $2^b + 2^c + 5 \equiv 5 \pmod{8}$, again not a square. It follows that $b$ equals either $1$ or $2$.

When $b = 1$, we look for $c$ such that $2^c + 7$ is a square. Obviously $c = 1$ works and it is the only possible value, since for $c \ge 2$ we have $2^c + 7 \equiv 3 \pmod{4}$.

Finally, when $b = 2$ we look for $c$ such that $2^c + 9 = k^2$, for some positive integer $k$. This can be written as $2^c = (k-3)(k+3)$, so we must have $k-3 = 2^\alpha$, $k+3 = 2^\beta$, where $\alpha < \beta$ are positive integers. Subtracting yields $2^\alpha(2^{\beta-\alpha} - 1) = 6$, which implies $\alpha = 1$ and then $c = 4$.

We conclude that the solutions are $a = b = c = 1$ and $\{a, b, c\} = \{1, 2, 4\}$.