Transistor with different voltage in colector and base

What you are seeing makes sense. Within nothing connected to the collector, the transistor acts like a diode from B to E. Current flows out the digital ouput, thru the 1 kΩ resistor, thru the transistor from B to E, thru the 470 Ω resistor, thru the LED, and back to the digital board. The LED is dimly lit because it effectively has 1.47 kΩ and the B-E voltage drop in series with it and a 5V supply. That probably results in around 1.5 mA LED current, which is visible but dim.

When 9V is applied to the collector, then the transistor can amplify. The collector current will be the base current times its gain. Let's say the LED drops 2 V and the B-E junction 700 mV. That leaves 2.3 V accross the 470 Ω (use designators already!) resistor for about 5 mA LED current. If the transistor has a gain of 50, then 1 part of this comes from the base and 50 parts of this come from the collector. The base current is then only about 96 µA, which causes only 96 mV drop accross the base resistor.

There is no need for the base resistor in your setup with the LED and its resistor in the emitter leg. Just get rid of it and tie the digital output directly to the base.

The base-emitter junction (BE) of a NPN transistor behaves like a diode:

So you basically have this equivalent circuit:

As you can see, your LED is in series with the resistor and the transistor's diode (the BE junction). Some current (depending on the value of your resistor) is passing through this circuit, making the LED light.

One way to avoid this is, as The Photon said, using a MOSFET. Another way is using the NPN transistor in the common-emitter configuration:

From this comment of yours:

Actually, I read the specifications on the transistor and the maximum base input is 5V

I can understand why you think you should drive the transistor's base from 4.5V (< 5V). But you can drive the base with a higher voltage as long as you have a resistor in series. The BE junction, acting as a diode, will have a fixed voltage drop (typically 0.7V) across it. The rest of the voltage will go across the resistor which also limits the current through the BE junction.