<s>[INST] Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
 [/INST] Let's grind it out. We want all ordered quadruples, but to get them it is enough to find all ordered triples $(a,b,c)$ where $a\le b\le c$. Then any triple of this kind we find can be permuted arbitrarily.
There cannot be triples $(a,b,c)$ with $a\le b\le c$ and $a\gt 2$, since for these we get that $3$ divides $a!+b!+c!$. So $a=1$ or $a=2$.
You don't seem to have considered $a=1$. So here goes. If $a=1$ then $b=c=1$ doesn't work. Let $b=1$. Then $c=2$ works, and we get the triple $(1,1,2)$. We also have the triple $(1,1,3)$. There is no triple of the shape $(1,1,k)$ for $k\gt 3$, for then $k!$ is divisible by $4$, so $1!+1!+k!$ cannot be a power of $2$.
Now look for triples of the shape $(1,b,c)$ where $b\ge 2$. There are none, for then $1!+b!+c!$ is odd.
Now look for triples of the form $(2,b,c)$. Try for $b=2$. Note that $c=2,3$  don't work, and $c\gt 3$ doesn't work, for in that case $8$ divides $c!$, so $2!+2!+c!$ cannot be a power of $2$.
So try for $b=3$. For $c$, calculate. We find, as you did, that $c=4$ and $c=5$ work. Nothing bigger than $5$ works, for if $c\gt 5$ then $c!$ is divisible by $16$, so $2!+3!+c!$ cannot be a power of $2$.
Now look at $b\ge 4$. This cannot work for any $c$, because then $2!+b!+c!$  has remainder $2$ on division by $4$.
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