Let $a=b$. Then $a(c-1)$ and $a^2-c=2^n$ are powers of 2, hence, $a=2^k,\,c-1=2^m$. Considering equality $2^{2k}=a^2=1+(c-1)+(a^2-c)=1+2^m+2^n$ modulo 4, we get $(m, \,n)=(0,\,1)$, $k=1$. This leads to solutions $(2,\,2,\,2)$ and $(2,\,2,\,3)$.
Now, assume that $a<b<c$. Denote $bc-a=2^A,\,ac-b=2^B$, then $2^A-2^B=(c+1)(b-a)>0$ hence $A>B$ and $2^B$ divides $(c+1)(b-a)$. Also, $2^B$ divides $2^A+2^B=(c-1)(a+b)$. Either $c-1$ or $c+1$ is not divisible by 4, hence either $2(b-a)$ or $2(b+1)$ is divisible by $2^B$, in both cases we have $a+b\ge 2^{B-1}$. So, $a(b+1)\le ac=2^B+b\le 2(a+b)+b$, i.e. $ab\le a+3b<4b$. Hence, $a<4$.
If $a=2$, then $2^C=2b-c,\,2^B=2c-b>1$, hence $B>0$. We get $b=\dfrac{(2^{C+1}+2^B)}{3},\,c=\dfrac{(2^{B+1}+2^C)}{3}$. If $C\ge 1$ then both $b$ and $c$ are even, hence $bc-2$ is not divisible by 4, in this case, $bc-2\ge 3\cdot 4-2=10$ cannot be a power of 2.
If $C=0$, then by modulo 3 we see that $B$ is even and $bc-2=\dfrac{(2+2^B)(2^{B+1}+1)-18}{9}=\dfrac{2^{2B+1}+5\cdot2^B-16}{9}$ must be a power of 2. This is not possible if $B\ge 6$, since 32 does not divide numerator $2^{2B+1}+5\cdot 2^B-16$ and it is greater than $16\cdot 9$. For $B=2$, we get $b=a=2$ and for $B=4$, we get $b=6$ and $c=11$.
If $a=3$, then $2^C=3b-c,\,2^B=3c-b\ge 2c+1>4$, hence $B\ge 3$. We get $b=\dfrac{(3\cdot 2^{C}+2^B)}{8},\,c=\dfrac{(3\cdot 2^{B}+2^C)}{8}$. We get $B>C$ (from $b<c$) and since $B\ge 3$, we get that $C\ge 3$ as well and $2^A=bc-3=(3\cdot 2^{C-3}+2^{B-3})(3\cdot 2^{B-3}+2^{C-3})-3$. If $B>C>3$, this is odd and greater than 1, hence they are not power of 2.
If $C=3$ we get $2^A=2^{B-3}(10+3\cdot2^{B-3})$. This is a power of 2 if and only if $B-3=1 \implies B=4$, $b=5,\,c=7$.
Hence, $(a,\,b,\,c)=(2,\, 2,\, 2),\,(2,\, 2,\, 3),\,(2,\, 6,\, 11),\,(3,\, 5,\, 7)$ and their permutations.
