How many distinct ways to climb stairs in 1 or 2 steps at a time?

Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.

The solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach.

Here is an illustration of what you are trying to solve for the case of $n=4$ steps (taken from this website, which also gives a combinatorial solution)

Any staircase with $n$ steps allowing paths with increments of 1 or 2 steps at a time will end up in one of two states before the last path is taken: either we've climbed $(n-1)$ steps already and have $\color{red}{one}$ more step to take, or we've climbed $(n-2)$ steps already and we have $\color{blue}{two}$ more steps to take (if we took only one step here then we'd end up in an arrangement from the first state).

Thus, to get the total number of possible ways to climb $n$ steps, we just add the number of possible ways we can climb $(n-1)$ steps and the number of possible ways we can climb $(n-2)$ steps, giving the familiar recurrence relation:

\begin{equation*} F_n = \left\{ \begin{array}{l@{}l@{}l} 1 & n = 0,1\\ \color{red}{F_{n-1}} + \color{blue}{F_{n-2}} & n \ge 2 \end{array} \right. \end{equation*}