As the title suggests, this is a discrete random variable with a countably infinite number of possibilities. I am trying to find the number of coins flips it takes to get a least 1 heads and 1 tails. So my first flip is either heads or tails, then my second flip is heads or tails, and so on. So my sample space would be {(H,T),(T,H),(T,T,H),(H,H,T)...} So it looks there is 1/2 chance that my experiment ends in 2 flips, 1/4 it ends in 3 flips, 1/8 it ends in 4 flips, and so on. So my PMF would seem to take the form of $p(x)={1/2^{n-1} , n=2,3,4,5,...}$ and 0 otherwise.
So the expected value would be $E[X]=\sum_{n=2}^{\infty}n{\frac1{2^{n-1}}}$
So to solve this I let m=n-1 to get $E[X]=\sum_{m=1}^{\infty}(m+1){\frac1{2^{m}}}$
I feel like this should equal 3 but I am not sure exactly how to express that.
Continuing from your working:
\begin{align} \sum_{m=1}^\infty \frac{m+1}{2^m} &= \sum_{m=1}^\infty \frac{m}{2^m} + \sum_{m=1}^\infty\frac{1}{2^m} \\ &=\sum_{m=1}^\infty m (0.5)^{m-1}(0.5) + \frac12 \cdot \frac{1}{1-0.5}\\ &= \frac1{0.5}+\frac12 \cdot \frac{1}{0.5} \\ &= 3 \end{align}
The first term is due to that is the expression of the expected value of geometric distribution with success probability $0.5$ and the second term is just a geometric sum.
A faster working is actually identifying it as $$1+X$$
where $X$ is a geometric distribution with success probability $0.5$. $1$ comes from the first toss and the success event is to get a different outcome from the first toss. From there you can obtain the expected value, pmf, and the variance.
We have $E[1+X]=E[X]+1$ and $Var(X+1)=Var(X)$.