$n^{2}+20n+15=n^{2}+2.10.n+100-85=(n+10)^{2}-85$
Let,
$(n+10)^{2}-85=m^{2}$
or,$(n+10)^{2}-m^{2}=85$
or, $ (n+10+m)(n+10-m)=85$
we can write ,
$85=17*5 ; 85=1*85 ;85=(-1)*(-85); 85=(-17)*(-5)$
So, let try with first one $85=17*5$ ,
$n+10+m=17$
$n+10-m=5$
So, $2n+20=22$
or, $n=1$
So , let see $n=1$ gives a perfect square or not?
$1^{2}+1*20+15=36=6^{2}$
Now let try with , $85=(-17)(-5)$
$n+10+m=-17$
$n+10-m=-5$
so, $2n+20=-22$
so, $n=-21$
So , let see $n=-21$ gives a perfect square or not?
=>$(-21)^{2}+(-21)*20+15=36=6^{2}$
Now let try with , $85=(85)(1)$
$n+10+m=85$
$n+10-m=1$
so, $2n+20=86$
so, $n=33$
So , let see $n=33$ gives a perfect square or not?
=>$(33)^{2}+(33)*20+15=1764=42^{2}$
Now let try with , $85=(-85)(-1)$
$n+10+m=-85$
$n+10-m=-1$
so, $2n+20=-86$
so, $n=-53$
So , let see $n=33$ gives a perfect square or not?
=>$(-53)^{2}+(-53)*20+15=1764=42^{2}$
So, The Here 4 solutions are possibles which are $1,-21,33,-53$ .
Check this comment here for some more insight .