Intensity, in a manner of speaking, is how bright the source of light is. Brighter being more intense.
What Einstein postulated (inspired by Planck) was that light was made up of packets of energy $\epsilon$ (proportional to frequency but we’ll get to that later). So Einstein said if the total energy of a monochromatic light source that you observe per unit area per unit time (intensity) is $E$, then it is made up of $n$ photons. Where n is given by:
Note that $\epsilon$ is a property of individual packets of light where as intensity $E$ is only defined for a collection of such packets.
Each packet can knock out one electron from the metal provided it has energy $\epsilon$ greater than the binding energy of the electron historically known as the work function $\phi_0$. The ejected electrons then have the kinetic energy $\epsilon - \phi_0$
Now if we were to increase the brightness of our source, we would be increasing the intensity $E$. Since $\epsilon$ is still the same (as it is the same source), our equation above says that we’ll have a bigger $n$. This means more electrons will be ejected.
Now coming to your question:
I = nhf
In both the equations we see that intensity and frequency are related. But in Einstein's theory both have different results.
It is further seen that the individual energy packets $\epsilon$ are related to frequency as $\epsilon=hf$.
Now, say we have a light source with intensity $E=100$ and the work function $\phi_0=1$. Consider the following two cases:
Both cases there’ll be ejection of electrons. However the number or electrons ejected aren’t the same. In the first case there will be $n=10$ electrons each with kinetic energy of $9$ units. In the second case, only $n=2$ electrons will be ejected albeit with a much higher kinetic energy of $49$ units.
Now let us double the brightness in the same experimental setup. So we have $E=200$ and $\phi_0=1$. Same two cases of:
Again in both cases there’ll be ejection of electrons. In the first case there will be $n=20$ electrons each with kinetic energy of $9$ units, same as before. In the second case, only $n=4$ electrons will be ejected albeit with a much higher kinetic energy of $49$ units, same as before.
What we observe is that the kinetic energy of each ejected electron is unchanged with changing intensity. However the number of ejected electrons has doubled with doubling intensity.