I'll prove that this holds in general, not just for 133
Let $P_i$ be the solution to this problem for number $i$. That is, the answer to the exact question is $P_{133}$
Let $F_i$ be the $ith$ Fibonacci number.
Proof by induction that $\forall i\geq1:P_i = F_{i+1}$
Base:
$i=1$ Can make with only a single 1. So $P_1 = 1 = F_2$
Hypothesis:
Assume holds for $i=k$ and $i=k-1,k>1$
Step:
Show that $P_k = F_{k+1} \rightarrow P_{k+1} = F_{k+2}$
From all solutions to $P_k$, we can add 1 and it is a solution to $P_{k+1}$.
From all solutions to $P_{k-1}$, we can add 2 and it is a solution to $P_{k+1}$
This is an exhaustive list of ways to make solutions to $P_{k+1}$, as going from any lower number, addings 1's and 2's will pass through either $P_k$ or $P_{k-1}$
Therefore
$P_{k+1} = P_k + P_{k-1} = F_{k+1} + F_k = F_{k+2}$ By hypothesis
(This is a bit of a crude proof. Mainly because I refer to $P_i$ as both a number and a set of solutions. But I think it gets the point accross)