7. Given the sequence $\left\{a_{n}\right\}$:
$$
a_{n}=2^{n}+3^{n}+6^{n}+1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Does there exist an integer $k \geqslant 2$, such that $k$ is coprime with all numbers in the sequence $\left\{a_{n}\right\}$? If it exists, find the smallest integer $k$; if not, explain the reason. (Zhang Lei provided the problem)
7. The smallest positive integer that satisfies the condition is 23.

First, the smallest integer that satisfies the condition must be a prime number, so it is only necessary to find the smallest prime that satisfies the condition.
By Fermat's Little Theorem, it is easy to conclude:
If a prime $p \mid\left(2^{n}+3^{n}+6^{n}+1\right)$, then $p \mid\left(2^{r}+3^{r}+6^{r}+1\right)$,

where $r(0 \leqslant r \leqslant p-2)$ is the remainder of $n$ divided by $p-1$.
From the conclusion, we know that if
$$
p \nmid\left(2^{r}+3^{r}+6^{r}+1\right)(r=0,1, \cdots, p-2) \text {, }
$$

then $p \nmid\left(2^{n}+3^{n}+6^{n}+1\right)$;
and if a prime $p \mid\left(2^{r}+3^{r}+6^{r}+1\right)$, then
$$
p \mid 6^{p-1-r}\left(2^{r}+3^{r}+6^{r}+1\right),
$$

which means $p \mid\left(3^{p-1-r} \times 2^{p-1}+2^{p-1-r} \times 3^{p-1}+6^{p-1}+6^{p-1-r}\right)$.
By $2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p)$,
$6^{p-1} \equiv 1(\bmod p)$,
then $p \mid\left(3^{p-1-r}+2^{p-1-r}+1+6^{p-1-r}\right)$.
Therefore, it is only necessary to verify $r=1,2, \cdots,\left[\frac{p}{2}\right]$.
When $p=2,3$, $p \mid(2+3+6+1)$;
When $p=5$, $p \mid\left(2^{2}+3^{2}+6^{2}+1\right)$;

When $p=7$, $p \mid\left(2^{3}+3^{3}+6^{3}+1\right)$;
When $p=13,17,19$, in all cases,
$$
p \left\lvert\,\left(2^{\frac{\nu-1}{2}}+3^{\frac{e-1}{2}}+6^{\frac{n-1}{2}}+1\right)\right. \text {; }
$$

When $p=23$, it is verified that
$23 \nmid\left(2^{i}+3^{i}+6^{i}+1\right)(i=0,1, \cdots, 11)$.
Thus, the smallest positive integer that satisfies the condition is 23.