Let $x_{1},x_{2}$ be two roots of the equation $$x-\ln{x}-m=0\; (m>1).$$ Show that $$1-\ln{m}<x_{1}x_{2}<\dfrac{\ln{m}}{m-1}.$$
So far, I only made $$\begin{cases} x_{1}-\ln{x_{1}}=m\\ x_{2}-\ln{x_{2}}=m \end{cases}\Longrightarrow x_{1}-x_{2}=\ln{\dfrac{x_{1}}{x_{2}}}$$ and set $\dfrac{x_{1}}{x_{2}}=t>1$, then we have $$x_{2}=\dfrac{\ln{t}}{t-1},x_{1}=\dfrac{t\ln{t}}{t-1}.$$ So it remains to prove $$1-\ln{m}<\dfrac{t\ln^2{t}}{(t-1)^2}<\dfrac{\ln{m}}{m-1}.$$
The following discussion follows the approach to express the two inequalities in question as dependent on just one single parameter $s = \frac12 \ln t$, where $t$ has already been introduced by the OP as $\dfrac{x_{1}}{x_{2}}=t>1$. Then one can write the two inequalities as $f(s) > 0$ and $g(s) > 0$ and show these.
Let's call the product $p = x_1 x_2$. As alread posted, $ p =\dfrac{t\ln^2{t}}{(t-1)^2}$, which can be rewritten $ p =\dfrac{s^2}{((\sqrt t-1/\sqrt t )/2)^2} = \dfrac{s^2}{(\sinh(s))^2}$.
Further, consider the general equality $4 x_1 x_2 = (x_1+x_2)^2 - (x_1-x_2)^2$. By the equation in question, $4 x_1 x_2 = (2m + \ln(x_1 x_2))^2 - (\ln(x_1/x_2))^2$ or $$ p = (m + \frac12 \ln(p))^2 - s^2 $$ This gives $m = - \frac12 \ln(p) + \sqrt{p +s^2}$, and, inserting $p$ from above, $m = - \frac12 \ln(\dfrac{s^2}{(\sinh(s))^2}) + \sqrt{\dfrac{s^2}{(\sinh(s))^2} +s^2}$, or simplified,
$$ m = - \ln(s) + \ln(\sinh(s)) + s \coth(s) $$
One observes that as $s=0$, $m(0) = 1$, and $m$ is increasing monotonically and unboundedly with $s$. So we have to consider the positive interval of $s$.
For the left bound we need to establish $f = p - 1 + \ln(m) > 0$. Since $p>0$ anyway, this needs only be considered for $1\leq m < $e . Since at $s= 2.1$, $m(s) >$e, we consider the range $0\leq s<2.1$
Inserting $p$ and $m$ gives
$$ f(s) = \dfrac{s^2}{(\sinh(s))^2} - 1 + \ln(- \ln(s) + \ln(\sinh(s)) + s \coth(s)) $$
This looks as follows:
Further, near $s=0$, by expansion, $f \simeq s^2/6 \geq 0$. This establishes the left inequality.
For the right inequality, we need to establish $g = \frac{ \ln(m) }{m-1} - p> 0$. Inserting $p$ and $m$ gives
$$ g(s) = - \dfrac{s^2}{(\sinh(s))^2} + \frac{\ln(- \ln(s) + \ln(\sinh(s)) + s \coth(s))}{- \ln(s) + \ln(\sinh(s)) + s \coth(s) - 1} $$
Near $s=0$, by expansion, $g \simeq s^2/12 \geq 0$.
The overall behaviour looks as follows:
For larger $s$, we have that $g(s)$ decreases monotonically with the limit $g(s\to \infty) = 0$.
This establishes the right inequality.
$\Box$