WLOG , due to symmetry of $a,b$ we can assume $a\geq b$ . $\textbf{Case 1:}$ $a>b$ then $b!|a!+b!=2^{c!}$ Now this follows that $b<3$ cause $b!$ has a prime divisor greater than $2$ if $b$ is greater than equal to $3$ ,thus $b=1,2$ But $a$ is atleast $1$ and $a,b$ are distinct so , if $b=1$ then the problem is $a!+1=2^{c!}$ now ,If $a>b=1$ thus $a!$ is even thus $a!+1$ is odd which is not possible for any $c$ is nonneagitive integer. Now if $b=2$, then $a!=2(2^{c!-1}-1)$ Then clearly $a<4$ because $4$ doesnot divide $a!$ thus if $a>b=2$ implying $a=3$ but then $3=2^{c!-1}-1$ implying $c!-1=2$ Thus $c!=3$ This gives no solution as $c!=3$ has no solution. Thus the problem is may possible if ,$a=b$ but not from this case. (Keep in Mind that we are not allowing zero,we will study it after we solve for problem in positive integers,the same is on remark later). $\textbf{Case 2:}$ $a=b$ Thus $a!=2^{c!-1}$ thus $a<3$ hence $a=1$ is a solution which indeed works with $a=b=1$ and $c!-1=0$ implying $c=1$ also ,$a=b=2$ then $c!-1=1$ implying $c=2$. Thus solutions are, $(a,b,c)=(1,1,1),(2,2,2)$ $\textbf{Remark:}$ Cases Like $a=0$ or $b=0$ or $c=0$ we can do the same thing but here I am taking Natural numbers are $>0$. For example :If $b=0$ $a!+1=2^{c!}$ , which was already done before in Case 1. Expect for the fact that we had taken $b>0$ in Case 1. Second thing is if $c=0$ then $a!+b!=2$ In other words ,If take non negative numbers instead , Then all solutions $(a,b,c)$ are $(1,1,1),(1,1,0),(1,0,0),(1,0,1),(0,0,0),(0,0,1),(0,1,1),(0,1,0),(2,2,2)$