The problem as in the title. What I tried so far: I investigated a function $f(x) = \frac{x^2}{(1+10^{-3})^x}$ for $x \geq 0$
So what I found out is that it's maximum is $x = \frac{2}{ln(1+10^{-3})}$ and it is monotoneously increasing/decreasing as $x$ goes from $0$ to $\frac{2}{ln(1+10^{-3})}$ \ from $\frac{2}{ln(1+10^{-3})}$ to $+ \infty$ So the desired $k$ is be either $\left \lfloor{\frac{2}{ln(1+10^{-3})}} \right \rfloor$ or $\left \lceil{\frac{2}{ln(1+10^{-3})}}\right \rceil $
But I don't think it counts as a solution of the problem, so I wonder if there is other way.
Your calculus method is fine. As an alternative, let $f(k)$ be our function. Then $$\frac{f(k+1)}{f(k)}=\left(1+\frac{1}{k}\right)^2\cdot \frac{1}{1+t},\tag{1}$$ where $t=10^{-3}$. So for a while $f$ is increasing. The largest $f(k)$ is reached for the smallest $k$ for which the ratio in (1) is less than $1$.
This is the smallest $k$ such that $1+\frac{1}{k}\gt \sqrt{1+t}$. The rest can be calculator work. It can also be done without a calculator, since $\sqrt{1+t}$ is very close to $1+\frac{t}{2}$, but a tiny bit less. .