Inspired by a Russian problem

$\gcd(7,x)=1$ gives the hint to check for modulo 7.

Since $7|2016$ but $7$ does not divides $x$ ,this implies $\gcd(7,y!)=1$ .

This can only happen if $y\le 6$ ,

If we observe that $44^{2}<2016 \rightarrow 2016+y!\ge 45^{2}$ , this gives that $y\ge4$

We check for the cases $y=4,5,6$ we conclude that none of the values give an integer solution for $x$ .

Taking mod 7 is all it takes.