As it says in the title I want to show that $\mathbb Z[\frac{2+i}{5}]\cap \mathbb Q =\mathbb Z$.
Set $\omega=\frac{2+i}{5}$. $\mathbb Z[\omega]$ is the smallest ring that contains $\mathbb Z$ and $\omega$. Therefore,
\begin{equation} \mathbb Z[\omega] = \{a_0+a_1 \omega+\cdots+a_k \omega^k\mid a_i\in \mathbb Z,k\in \mathbb N\} \end{equation}
Let, then, $q\in\mathbb Z[\frac{2+i}{5}]\cap \mathbb Q$. Since $q\in\mathbb Q$ we can write $q=\frac{m}{n}$, where $m,n$ are relatively prime integers and $m\neq 0$. Now, since $q\in\mathbb Z[\frac{2+i}{5}]$ we have that
\begin{equation} q=a_0+a_1 \omega+\cdots+a_k \omega^k \end{equation}
for some $a_i\in \mathbb Z$. Then,
\begin{align} &\frac{m}{n}=a_0+a_1 \frac{2+i}{5}+\cdots+a_k \left(\frac{2+i}{5}\right)^k \\ \Longrightarrow \ &5^km=5^kna_0+5^{k-1}na_1(2+i)+\cdots+a_kn(2+i)^k \end{align}
Hence, $n\mid 5^km$ and since $\gcd(n,m)=1$ we have that $n|5^k$.
Thus, every rational number in $\mathbb Z[\omega]$ is of the form $\frac{m}{5^l}$ where $m\in\mathbb Z$, $l$ is a positive integer and when $l\geq 1$ it holds that $5\nmid m$. It suffice then to show that it always hold that $l=0$. I am stuck here. I would very much appreciate some help!
Hint: Look at the equation $5^km = n(5^ka_0+5^{k-1}a_1(2+i)+ \cdots + a_k(2+i)^k)$ as an equation between Gaussian integers. Note that $2+i$ is a prime in the ring of Gaussian integers arising from the factorization of the integer prime $5$, namely $(2+i)(2-i)=5$. You already noticed that $n$ is a power of $5$. In particular, if $m/n$ is not an integer, you may assume that $m$ is not divisible by $5$. Hence, viewing it as a Gaussian integer, it should not be divisible by $2+i$ or $2-i$. Can you see the contradiction from here?