Attempt:
I thought that by fixing one of the index and doing it conversely I can find accumulation points as follows:
For every fix $n_0\in \mathbb N$ $2^{-n_0}+5^{-m}$ converges to $2^{-n_0}$
and For every fix $m_0\in \mathbb N$ $2^{-n}+5^{-m_0}$ converges to $5^{-m_0}$
and letting free both indices we have a accumulation point $0$
So all accumulation points are $ {2^{-n},n\in\mathbb N^+ } \cup {5^{-n},m\in\mathbb N^+} \cup 0 $$ is it my reasoning right?
You are half-way through: You showed that each o these points is an accumulation point. The question remains whether any other point can be an accumulation point.
Let $x\in \Bbb R$ be a number that is an accumulation point, but not among the ones you found. Clearly $x\ge 0$, and as you already found that $0$ is an accumulation point, we have $x>0$. There are $n_0,m_0\in\Bbb N$ such that $2^{-n}<\frac13x$ for all $n>n_0$ and $5^{-m}<\frac13x$ for all $m>m_0$. Hence all $2^{-n}+5^{-m}$ that differ from $x$ by less than $\frac13x$ must have $n\le n_0$ and/or $m\le m_0$. Then for any sequence of elements of the given set that converges to $x$, each term falls into (at least) one of the finitely many categories $n=1$, $n=2$, $\ldots$, $n=n_0$, $m=1$, $m=2$, $\ldots$, $m=m_0$. Hence we find a subsequence such that all terms of it fall into the same category, that is, on of the cases with one exponent fixed that you already covered.