7. Given an arithmetic sequence $\left\{a_{n}\right\}$ with a positive integer common difference $d$, and a geometric sequence $\left\{b_{n}\right\}$, and $a_{1}=b_{1}$, $a_{2}=b_{2}$, the values of $d$ are $2,3,4,5$, the value of $d$ that can make all terms of $\left\{b_{n}\right\}$ also terms of $\left\{a_{n}\right\}$ is $\qquad$ .
7. $2,3,4,5$.

From the problem, we know $a_{n}=a_{1}+(n-1) d, b_{n}=a_{1} q^{n-1}$, then $a_{1}+d=a_{1} q$. For all terms of $\left\{b_{n}\right\}$ to be terms of $\left\{a_{n}\right\}$, it is necessary that for any $n \geqslant 3$, there exists $m \in \mathbf{N}_{+}$ such that $b_{n}=a_{m}$ holds, i.e., $a_{1} q^{n}=a_{1}+(m-1) d$. Thus, $m-1=\frac{a_{1}}{d}\left[q^{n}-1\right] \in \mathbf{N}_{+}$. This is satisfied as long as $a_{1}=d$, at which point $q=2, m=2^{n} \in \mathbf{N}_{+}$ holds. $d=2,3,4,5$ all meet the requirements.