Showing that $1+2020^y$ ($y\in \mathbb{Z}$) is not a perfect square for even $y$ it is obvious that $1+2020^y$ is not a perfect square since $2020^y$ is a perfect square $+1$.+
for uneven $y$, i wanted to show that $z^2-1=(z-1)(z+1)=2020^y$. How can I show that I can't distribute the factors of $2020^y$ into two integers $n$ and $n+2$?

To answer the question you asked: "How can I show that I can't distribute the factors of $2020^y$ into two integers n and n+2?"
$2020^y = 101^y*2^{2y}*5^y$.  
If $n= z-1, n+2 = z+1$ then $\gcd(z+1, z-1) = \gcd((z+1)-(z-1),(z-1))=\gcd(2,z-1)=\begin{cases}1\\2\end{cases}$
As $z-1, z+1$ are either both even or both odd, and $2020^y$ is even they are both even and $\gcd(z-1, z+1) =2$ and so $2$ divides one of $n$ or $n+2$ and $2^{2y-1}$ divides the other.
And $5^y$ divides one of $n$ or $n+2$ and $101^y$ also divides one or the other.
And yet the difference between $n$ and $n+2$ must be $2$ which is, in the grand scale of things very close. $101^y$ is a very large factor that it pales all the others combined. Its reasonable to assume the distribution of $2, 2^{2y-1}, 5^y,101^y$ that has the smallest difference will be $2*101^y$ vs $2^{2y-1}*5^y$ and the difference is much more than $2$.
In other words:  $2*101^y - 2^{2y-1}*5^y = 2*101^y - 5*20^{y-1}=2*101^y-\frac{2*20^y}{8}>2(101^y-20^y)=2(101-20)(101^{y-1}+....+20^{y-1})\ge 2(81)> 2$.
(Assuming, of course, $y > 0$.  But if $z =0$ then $1+2020^0 = 2$ and if $z < 0$ than $1+2020^{z}$ is not an integer.)
ANd if $A-B > 2$ then $A*k - \frac Bk > A-B > 2$ for any $k> 1$.  And all difference between these distributions of factors will be $2^{m}5^j*101^y - 2^a*5^b=k(2*101^y) - \frac{2^{2y-1}*5^y}k $ where $m= 1$ or $2y-1$ and $j=0$ or $y$ and $a=2y-1$ or $1$ and $b=y$ or $0$.
......
But, all in all, the accepted answer is a much better approach for the problem.
