Let $\displaystyle P_P(n,\mu)=\frac{\mu^n}{n!}\,e^{-\mu}$ and $\displaystyle P_G(x,\mu,\sigma)=\frac{1}{\sqrt{2\pi}\sigma}\,\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\quad$ where $\sigma=\sqrt{\mu}$
At what values of $\mu$ is $P_P(n=\mu+z\sqrt{\mu})\approx P_G(x=\mu+z\sqrt{\mu})$, to within $10\%$, for $z=(1,3)$ (round $\mu+z\sqrt{\mu}$ to the nearest integer, and evaluate $P_G$ at the same value)?
BACKGROUND
This is for a graduate astrophysics class and the only lecture we got on statistics was the full derivation of the binomial and Poisson distribution (from the binomial), and the fact that the Gaussian distribution was the Poisson but for large $\mu$ (brief mention of Central Limit Theorem).
MY ATTEMPT
First, I tried plugging $\mu+z\sqrt{\mu}$ in each distribution to see what I get:
$\begin{aligned} &\begin{aligned} P_P(n=\mu+z\sqrt{\mu})&=\frac{\mu^{\mu+z\sqrt{\mu}}}{(\mu+z\sqrt{\mu})!} \,e^{-\mu} \\[1ex] &=\mu^{\mu+z\sqrt{\mu}}\cdot \frac{1}{\sqrt{2\pi(\mu+z\sqrt{\mu})}}\left(\frac{e}{\mu+z\sqrt{\mu}}\right)^{\mu+z\sqrt{\mu}}\cdot e^{-\mu} \qquad \text{(Stirling's approx.)} \\[1ex] &=\frac{1}{\sqrt{2\pi(\mu+z\sqrt{\mu})}}\left(\frac{\mu}{\mu+z\sqrt{\mu}}\right)^{\mu+z\sqrt{\mu}}\cdot e^{z\sqrt{\mu}} \end{aligned} \\[5ex] &\begin{aligned} P_G(x=\mu+z\sqrt{\mu})&=\frac{1}{\sqrt{2\pi\mu}}\,\exp\left(-\frac{\Big[(\mu+z\sqrt{\mu})-\mu\Big]^2}{2\mu}\right) \\[1ex] &=\frac{1}{\sqrt{2\pi\mu}}\,\exp\left(-\frac{z^2\mu}{2\mu}\right) \\[1ex] &=\frac{1}{\sqrt{2\pi\mu}}\,\exp\left(-\frac{z^2}{2}\right) \end{aligned} \end{aligned}$
And from there, I'm kinda stuck. I have no clue how I could isolate $\mu$ if I put them equal. Moreover, I don't understand how I could add the fact that it has to be "within $10\%$" and also I don't get how I'm suppose to "round $\mu+z\sqrt{\mu}$ to the nearest integer".