Let $a \geq b \geq c$ if $c \geq 2$ then $2^a+2^b+2^c+3=3 \pmod 4$ and can not be square So $c=1$ $2^a+2^b+5$ is square $2^a+2^b \equiv 4 \pmod{8}$ so $(a,b)=(1,1)$ or $b=2,a \geq 3$ $2^a+9=n^2$ $n-3=2^u,n+3=2^v \to n=5,a=4$ So $(a,b,c)=(1,1,1),(4,2,1)$

WLOG let $a\le b\le c$. If $a\ge 2$ we have $2^a+2^b+2^c+3 \equiv 3(mod4)$ which is not a perfect square so $0\le a \le 1$. Case $a=0$ $\implies$ $ 2^b+2^c+4=r^2$ for some natural number $r$ and notice $r$ is even so $r=2k$ $\implies$ $ 2^b+2^c+4=4k^2$ $\implies$ $ 2^{b-2}+2^{c-2}+1=k^2$ and factoring we will have $2^i(2^{b-i-2}+2^{c-i-2})=(k-1)(k+1)$. Now notice $k+1>k-1$ and $c\ge b> i \ge 2$ so paring the factors from both sides we have $k-1=2^i$ $\implies k=2^i+1$ and looking at the second equation $2^{b-i-2}+2^{c-i-2}=k+1$ is equivalent to $2^{b-i-2}+2^{c-i-2}=2^i +2$ and dividing both sides by $2$ , $2^{b-i-3}+2^{c-i-3}=2^{i-1}+1$ but notice that if $i\ge 2$ then the $RHS$ will be odd and we know that $i\ge 2$ so the only possible case is where all the exponents from the powers of $2$ are equal to $0$ therefore: $b-i-3=0$ $\implies b=i+3$ and $c=2i+2$ . Case $a=1$ We get $2^b+2^c+5$ is a perfect square and notice $2^1+2^1+5=9$ works and if $b,c\ge 3$ then $2^b+2^c+5 \equiv 5 (mod8)$ which is not a square so $b=2$ $\implies$ $ 2^b+9=z^2$ for some natural number $z$ then $(z-3)(z+3)=2^b$ . $z-3=2^s$ and $z+3=2^p$ and isolating $z$ you will get $a=4$. Finally we have all the triples $(a,b,c)=(1,1,1),(1,2,4), (0, i+3, 2i+2)$ with $a\le b\le c$ and $i\ge 2$ where $i$ is natural number.