Assuming I understand the situation correctly:
the number of incidents for an individual is $Y_i \sim \text{Pois}(\lambda \, T_i)$ where the incidence rate per unit time is $\lambda$, $T_i$ is the exposure time for individual $i$ and the $Y_i$ variates are independent. Then we define the random variable $N=\sum_i Y_i$ and the total exposure $T=\sum_i T_i$, in which case $\hat{\lambda}=N/T$ is the MLE of $\lambda$
you seek a CI for $\hat{\lambda}$
$\text{Var}(\hat{\lambda}) = \text{Var}(\frac{N}{T}) = \frac{1}{T^2}\text{Var}(N)=\frac{1}{T^2}\text{Var}(\sum_iY_i)=\frac{1}{T^2}(\sum_i\lambda T_i)=\lambda/T$
Hence the estimated standard error is $\sqrt{\hat{\lambda}/T} = \frac{\sqrt{N}}{T} = \frac{N\times \frac{1}{\sqrt{N}}}{T}=\frac{N}{T}\times \frac{1}{\sqrt{N}}=\frac{\text{IR}}{\sqrt{N}}$
So (if $N>0$), either of these two formulas is okay:
$\text{s.e.}(\text{IR}) = \sqrt{\frac{IR}{T}} = \frac{IR}{\sqrt{N}}$
and so a large sample (large-$N$) interval for the $\text{IR}$ would multiply that by a suitable $Z$ value to get the interval half-width either side of the estimate.
You seem to have perhaps muddled up the roles of $N$ and $T$ in your calculation
Actually, it looks like the link you gave explained how their formula came about pretty clearly in the couple of lines under "If $N$ is large".
