Suppose $\mu$ and $\nu$ are two probability measures on $[0,1]$. Let their $n$-th moments be denoted by $\mu_n$ and $\nu_n$, respectively, for $n \in \mathbb{N}$.
If we know that $\mu_n=\nu_n$ for infinitely many $n$, can we conclude that $\mu=\nu$?
One way to resolve this would be to see if the span of $\{ x^n \mid n \in S \}$ with $|S|=\infty$ is dense in the set of continuous functions $C[0,1]$. Is such a set always dense?
We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.
Recall Müntz-Szász theorem, which states (in particular) the following:
Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:
the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.
If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.
If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.
If you are willing to move one dimension higher, then having infinitely many equal moments becomes rather weak condition. There are examples of uniform measures supported on finite polygons that have all their harmonic (a.k.a. complex) moments equal, yet the supports differ. See e.g. page 2 here.
