You're working with an I2C bus, right? I'm going to call the lines on the 3.3 V side SDA3 and SCL3; the two lines on the 5.0 V side SDA5 and SCL5.
"MOSFET and two resistors"
As todbot and cyphunk already pointed out, the "MOSFET and two resistors" circuit described in the SparkFun "sensor interfacing" tutorial does what you want:
The logical data flow is symmetric -- data flows in both directions on the SDA lines, from the low side to the high side and, milliseconds later, from the high side to the low side.
The "will not work the other direction" note points out the circuit is physically asymmetric: your 5.0 V I²C device must be connected to the "high side", your 3.3 V device must be connected to the "low side". Because the circuit is physically asymmetric, it's certainly not obvious that it is logically symmetric.
(That tutorial links to a AN97055 application note that shows a "two MOSFETs and two resistors" circuit that is physically symmetric, and therefore is obviously logically symmetric).
The lines labeled "TX" on the SparkFun converters -- that the original poster pointed out -- implement that "MOSFET and two resistors" bidirectional circuit.
So connect SDA3 to TX_LV, SDA5 to TX_HV, SCL3 to TX2_LV, and SCL5 to TX2_HV.
Then data flows in both directions: When your low-side device drives the SDA3 and SCL3 pins, the appropriate voltages are seen on the high-side SDA5 and SCL5 pins.
Milliseconds later, when the high-side device drives the SDA5 and SCL5 pins, the appropriate voltages are seen on the SDA3 and SCL3 pins.
(Inconsistently, the lines labeled "RX" on that converter board will only transmit data in the high-voltage to low-voltage direction.)
Bi-directional optoisolator
Since you are using I²C, you might also be interested in a bi-directional optoisolator for I²C. The two-optoisolator circuit is more expensive and slower than the "MOSFET and two resistors" circuit, but it works when one side has signals that swing between 0 V and 5.0 V, and the other side has signals that swing between 500.0 V and 505.0 V.
The two-optoisolator circuit is also completely physically symmetric -- and therefore logically symmetric -- it doesn't matter which side is the high side and which side is the low side.