jmc

We have to use a little bit of casework to solve this problem because some numbers on the die have a positive difference of $2$ when paired with either of two other numbers (for example, $3$ with either $1$ or $5$) while other numbers will only have a positive difference of $2$ when paired with one particular number (for example, $2$ with $4$).

If the first roll is a $1,$ $2,$ $5,$ or $6,$ there is only one second roll in each case that will satisfy the given condition, so there are $4$ combinations of rolls that result in two integers with a positive difference of $2$ in this case. If, however, the first roll is a $3$ or a $4,$ in each case there will be two rolls that satisfy the given condition- $1$ or $5$ and $2$ or $6,$ respectively. This gives us another $4$ successful combinations for a total of $8.$

Since there are $6$ possible outcomes when a die is rolled, there are a total of $6\cdot 6=36$ possible combinations for two rolls, which means our probability is $\frac{8}{36}=\boxed{\frac{2}{9}}.$

OR

We can also solve this problem by listing all the ways in which the two rolls have a positive difference of $2:$ $$(6,4),(5,3),(4,2),(3,1),(4,6),(3,5),(2,4),(1,3).$$ So, we have $8$ successful outcomes out of $6\cdot 6=36$ possibilities, which produces a probability of $8/36=2/9.$