Hint: You can treat this as a quadratic equation, if you think of $m$ as your variable and $n$ as a constant:
$$
\underbrace{m^2}_{\text{quadratic}}+\underbrace{(-2n-1)m}_{\text{linear}}+\underbrace{n^2}_{\text{constant}}=0
$$
In other words, your quadratic is $am^2+bm+c$, where $a=1$, $b=(-2n-1)$, and $c=n^2$. With this, you can come up with two possibilities for how we can write $m$ in terms of $n$. Then, it is a matter of exploring the constraints.
Alternatively:
Since we require that $m>n$, we know that $m-n>0$. We also know that $m>0$. Our equation can be rewritten as $(m-n)^2-m=0$, or $(m-n)^2=m$; since $m>0$, we can take square roots of both sides to get
$$
\lvert m-n\rvert=\sqrt{m}.
$$
What simplifications can be made here based on the assumption that $m-n>0$? What does this lead you to?
Hint: You can treat this as a quadratic equation, if you think of $m$ as your variable and $n$ as a constant: $$ \underbrace{m^2}_{\text{quadratic}}+\underbrace{(-2n-1)m}_{\text{linear}}+\underbrace{n^2}_{\text{constant}}=0 $$ In other words, your quadratic is $am^2+bm+c$, where $a=1$, $b=(-2n-1)$, and $c=n^2$. With this, you can come up with two possibilities for how we can write $m$ in terms of $n$. Then, it is a matter of exploring the constraints.
Alternatively:
Since we require that $m>n$, we know that $m-n>0$. We also know that $m>0$. Our equation can be rewritten as $(m-n)^2-m=0$, or $(m-n)^2=m$; since $m>0$, we can take square roots of both sides to get $$ \lvert m-n\rvert=\sqrt{m}. $$ What simplifications can be made here based on the assumption that $m-n>0$? What does this lead you to?