Find all positive integer triplets $(x,y,z)$ in $x!+y!=15\cdot 2^{z!}$ 
Find all positive integer triplets $(x,y,z)$ in $x!+y!=15\cdot 2^{z!}$

Attempt: If $x\geq y,$ Then $x!\geq y!$ or $x!+x!\geq x!+y! = 15\cdot 2^{z!}$
So $2x!\geq 15\cdot 2^{z!}\Rightarrow x!\geq 15\cdot 2^{z!-1}$
could some help me how to go further, Thanks
 A: WLOG, assume $x\leq y$. Then, $$x!+y! = x!\left(1+\frac{y!}{x!}\right)=15\cdot2^{z!}\implies x\leq 5$$ since $7$ and $9$ does not divide $15\cdot2^{z!}$. 
If $x = 5$, then $1+\frac{y!}{x!} = 2^{t}\implies y = 5$. This is because if $y\geq6$ then $1+\frac{y!}{x!} $ is odd. However, $z$ does not exist because in this case $2^{z!} = 16$.
If $x = 4$, then $1+\frac{y!}{x!} = 5\cdot 2^t$. However, if $y\geq 5$ then $1+\frac{y!}{x!} \equiv 1 (mod \ 5)$, so this case is impossible as well.
If $x=3$, then  $1+\frac{y!}{x!} = 5\cdot 2^t$. similar to above, we see the only possible case is $y=4$. In this case, $z!=1$, so $(x,y,z) = (3,4,1)$.
If $x=2$, then $1+\frac{y!}{x!} = 15\cdot 2^t$. However, all $y\leq 4$ do not work, and when $y\geq5$, it is the same as above. Hence, this case is impossible.
Finally, if $x=1$, then $1+y! = 15\cdot 2^t$. However, all $y\leq 4$ do not work, and when $y\geq5$, it is the same as above. Hence, this case is impossible.
To conclude, the only cases are $(x,y,z) = (3,4,1) $ or $(4,3,1)$.
