Here is an elementary approach. We want to rule out finding distinct $a,b,c,d$ in short interval (specifically $[n^2+1,n^2+2n]$) with $ad=bc.$ We may assume that $a$ is the smallest and that $b<c$. Then $a<b<c<d.$ I will show that $a+2\sqrt{a}+1\leq d$ so that if $n^2 \leq a$ then $(n+1)^2 \leq d.$

Claim: There are integers $u<v$ and $x<y$ with $$a,b,c,d=ux,uy,vx,vy.$$

Proof: Let $u=\gcd(a,b)$ so $a=ux$ and $b=uy$ with $\gcd(x,y)=1.$ Then $uxd=uyc$ so $xd=yc$ and thus (since $x$ and $y$ are co-prime) there is $v$ with $c=xv$ and $d=yv.$

**ASIDE** We might as well use $v=u+1$ and $y=x+1$ since $$a,b',c',d'=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ give $ad'=b'c'$ with $a <b',c',d'\leq d.$ I'll comment a bit more about this at the end.

So we want to show

if $ad=bc$ with $n^2<a<b<c<d$ then $d>(n+1)^2.$

From the claim above, $n^2<a=ux$ and $d\geq (u+1)(x+1)=ux+u+x+1.$ But given $ux=a>n^2,$ we know $u+x\geq 2\sqrt{a}>2n.$ Thus $ux+u+x+1>n^2+2n+1$ as desired.

Consider this problem: Given $a$, find $a<b<c<d$ with $d$ minimal such that $ad=bc.$ The work above shows that the solution is to have $$a,b,c,d=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ with $|u-x|$ minimal and that $d>a+2\sqrt{a}+1.$ If we allow $b=c$ then $$n^2\cdot (n+1)^2=(n^2+n)\cdot (n^2+n).$$ If we want $b < c$ then $(n^2-n)\cdot(n^2+n)=(n^2-1)\cdot(n^2).$ With $d-a\approx 2\sqrt{a}.$

If $ad=bc$ then $abcd$ is a perfect square. However this property is weaker and there are solutions such as $$a,b,c,d=2\cdot 120^2,3\cdot 98^2,30\cdot 31^2,5\cdot 76^2=$$ $$28800, 28812, 28830, 28880$$ with $d<a+\frac12\sqrt{a}$ and all four factors between $28561=169^2$ and $28900=170^2.$