We know that $2016 = 2^5 \times 3^2 \times 7$ .

Now, suppose $y \geq 8$ , we have

$x^2 = 2016 + y!$

$x^2 = 2016 + 8!k$ where $k$ is an interger.

$x^2 = 2^5 \times 3^2 \times 7 + 2^7l$ where $l$ is an integer.

$x^2 = 2^5(63 + 4l)$

Now,since the expression in the bracket is odd, the term on the right contains an odd power of 2, which is not possible. Therefore $y < 8$ . That leaves us 7 cases $1 \leq y \leq 7$ . Checking all these, we find the only two solutions $(84,7) , (-84,7)$ . Therefore the maximum possible value of $y - x = 7 - (-84) = \boxed{91}$

$x^2-y!=2016. ~~~\sqrt{2016}~<~45~~~and~is~ not~an~integer.~And~x>44.\\ So~y\neq~0.~~and ~NOTE~that~both~are~integers.\\ x^2-y!>0,~~~\therefore~x^2>y!. ~~~~For~n>3, n!>n^2. ~~~\implies~~y<x.\\ \therefore~y-x<0. ~ for ~x>0.~~~and~y-x>0~for~x<0.\\ So~maximum~ of ~~y - x~is~ when~x<0,~and~y~-~x~>~0.\\ x=\sqrt{ 2016-y!}.~So~\sqrt{ 2016-y!}~ must~ be~ an~ integer.\\ Only~x= \pm 84~and~ y=7 ~satisfy~this.\\ Since~x<0,~~so~y - x = 7 +84= \Large \color{#D61F06}{91}.$

\(x^2-y!=2016

x=2n

4(n^2-504)=y!

2*3*4*6(k^2-14)=y!

k^2-14=5*7

k=7

then x=84 or -84 and y=7

so max{(7-84),(7--84)}=7+84=91\)

As you can see that this is equation is satisfying only one set of value and that is x= +84 & -84 and y=7. i.e, (84)^2 - 7(factorial) = 2016 so, to find the maximum value of y-x, we to make use of these value and the maximum value of this expression. that can be achieved by taking x= -84 and y=7 so, 7- (-84) = 91. which is the required answer.