We have to find the integers $m$ s.t. there exists a positive integer $n<m$ satisfying the conditions $(1) \;\; mn$ is a perfect square. $(2) \;\; m-n$ is a prime. $(3) \;\; 1000 \leq m \leq 2021$. First of all, there are two positive integers $a$ and $b$ with $a$ squarefree s.t. $m = ab^2$. Hence by condition (1) there is a positive integer $c<b$ s.t. $n=ac^2$. Consequently by condition (2) there is a prime $p$ s.t. $p = m - n = ab^2 - ac^2 = a(b - c)(b + c) = p$, implying $(a,b-c,b+c) = (1,1,p)$, which means ${\textstyle (a,b,c) = (1,\frac{p+1}{2},\frac{p-1}{2})}$. Therefore ${\textstyle (m,n) = ((\frac{p+1}{2})^2, (\frac{p-1}{2})^2)}$, which combined with condition (3) give us ${\textstyle \sqrt{1000} \leq \frac{p+1}{2} \leq \sqrt{2021}}$, yielding ${\textstyle 32 \leq \frac{p+1}{2} \leq 44}$, i.e. $63 \leq p \leq 87$. Hence we obtain $p \in \{67,71,73,79,83\}$. Conclusion: There are exactly five integers satisfying conditions (1)-(3).

$mn$ is a perfect square $m-n$ is a prime number $1000<m<2021$ If exists a number $d$ such that $d$ divides $m$ and $d$ divides $n$, then $d$ divides $m-n$, but $m-n$ is prime. We conclude that $d=1$, so $m$ and $n$ are both perfect squares. So $m=a^2$, $n=b^2$, $m-n=a^2-b^2=(a+b)(a-b)$, $m-n$ is prime, so $a=b+1$ Now let's check the perfect squares between $1000$ and $2021$: $31^2<a^2<45^2$. Just now check now which consecutive numbers whose sum is a prime number: They are $(33,34), (35,36), (36,37) (39,40), (41,42)$. So the values of $m$ are: $34, 36, 37, 40, 42$

$\color{blue} \boxed{\textbf{SOLUTION}}$ Let, $m = sq^2, n=sr^2$ So, $p = m - n = sq^2 - sr^2 = s(q - r)(q+ r)$, $\implies (s,q-r,q+r) = (1,1,p)$ $\implies {\textstyle (s,q,r) = (1,\frac{p+1}{2},\frac{p-1}{2})}$. $\implies {\textstyle (m,n) = ((\frac{p+1}{2})^2, (\frac{p-1}{2})^2)}$, So We have, ${\textstyle \sqrt{1000} \leq \frac{p+1}{2} \leq \sqrt{2021}}$, $\implies {\textstyle 32 \leq \frac{p+1}{2} \leq 44}$, $\implies63 \leq p \leq 87$. $\color{red} \boxed{p \in \{67,71,73,79,83\}}$