$B_\lambda$ and $I_\lambda$ are different but related physical quantities.
$B_\lambda$ is spectral radiance. It is the power per unit projected area per unit wavelength per unit solid angle radiated by the blackbody. “Projected” means that the area is to be measured perpendicularly to the direction of observation.
$I_\lambda$ is spectral exitance. It is the power per unit area per unit wavelength radiated by the blackbody.
The relationship between them is thus that the exitance is the radiance, times the cosine of the angle between the radiating surface and the direction of emission (to account for the projected area), integrated over all possible solid angles into which the radiation can be emitted. This is the hemisphere outside the radiating surface. So
$$I_\lambda=\int_h B_\lambda \cos\theta\,d\Omega,$$
where $h$ is a hemisphere.
As you can see, $B_\lambda$ is independent of the angle of emission, so this becomes
If you work out this elementary integral, you will find that it is equal to $\pi$, explaining the “discrepancy” that you noticed. So
To summarize: $I_\lambda$ integrates over solid angles, $B_\lambda$ does not. It is a subtle but important difference.
You can also express these quantities in terms of other properties of the radiation, such as its frequency $\nu$ rather than its wavelength $\lambda$. The usual notation is $B_\nu$ and $I_\nu$. The relationship is
and similarly for $I$.
There is no good reason to put in a factor of -1. I suppose some people do this because if these quantities are expressed in terms of frequency rather than wavelength, the spectrum “goes the other way”.