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?

Assume $1\le a\le b\le c$. Then $2^n=a!(1+b\cdots(a+1)+c\cdots(a+1)$.
It follows that either $a=2$ or $a=1$. 
Consider the case $a=2$. Then one of the numbers $b\cdots(a+1)$ and $c\cdots(a+1)$ must be even, and the other odd, and since $b\cdots(a+1)$ divides $c\cdots(a+1)$ it follows that $b\cdots(a+1)$ must be odd, and $c\cdots(a+1)$ must be even. Hence $b=2$ or $b=3$, and $c\ge 4$. So we are looking for $c$ such that either $\displaystyle 2+\frac{c!}2$ is a power of $2$, or $\displaystyle 4+\frac{c!}2$ is a power of $2$. 
These are only ideas how to possibly start (seems right), I will think if I could add more details and a more definite answer. Edit: I added all details below. 
Consider the case $a=1$. Then $2^n=1+b!+c!$. It follows that $b=1$ and $\displaystyle 2^n=2(1+\frac{c!}2)$, so $\displaystyle 2^{n-1}=1+\frac{c!}2$. This works if (and only if) $c=2$ or $c=3$. So we get solutions $(a,b,c)=(1,1,2)$ with $1+1+2=4=2^2$, and $(a,b,c)=(1,1,3)$ with $1+1+6=8=2^3$. It looks like you missed these small integers solutions. 
Back to the case $a=2$. First let $b=2$. Then $\displaystyle 2^n=2(1+1+\frac{c!}2)$, so $\displaystyle 2^{n-2}=1+\frac{c!}4$. No solutions in this case.  
Finally, let $a=2$, $b=3$. Then $\displaystyle 2^n=2(1+3+\frac{c!}2)$, so $\displaystyle 2^{n-3}=1+\frac{c!}8$. Works if $c=4$ or $c=5$. Solutions $(a,b,c)=(2,3,4)$ when $2+6+24=32=2^5$, and $(a,b,c)=(2,3,5)$ when $2+6+120=128=2^7$ (these are the two solutions that you found). This finishes all possible cases, no other solutions. 
