Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 can be divided by $x^{2}+1$.

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 translated text:
Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 can be divided by $x^{2}+1$.

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【Analysis】More explicitly, the polynomial $(x+1)^{n}-1$ needs to satisfy the equivalent condition:
There exist integer-coefficient polynomials $P$ and $Q$ such that
$$
(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x) \text {. }
$$

Assume without loss of generality that
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {. }
$$

Then $R(x)=3 Q(x)$, where $P(x)$ is the quotient and $R(x)$ is the remainder with coefficients all being multiples of 3.
Now consider $R(\mathrm{i})$.
Since the coefficients of the odd and even powers of $R(x)$ are all divisible by 3, the real and imaginary parts of $R(\mathrm{i})$ are also divisible by 3.
By calculation, we get
$$
\begin{array}{l}
(1+\mathrm{i})^{2}=2 \mathrm{i},(1+\mathrm{i})^{4}=-4, \\
(1+\mathrm{i})^{6}=-8 \mathrm{i},(1+\mathrm{i})^{8}=16 .
\end{array}
$$

Thus, $(1+\mathrm{i})^{8}-1=15$ has a real part and an imaginary part both divisible by 3.

Since the even powers of $1+\mathrm{i}$ are either purely real or purely imaginary and their coefficients are not divisible by 3, multiplying by $1+\mathrm{i}$ still results in an imaginary part that is not divisible by 3.
Testing $n=8$, we have
$$
\begin{array}{l}
(x+1)^{8}-1 \\
=x^{8}+8 x^{7}+28 x^{6}+56 x^{5}+70 x^{4}+ \\
56 x^{3}+28 x^{2}+8 x \\
=x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x \\
\equiv\left(x^{2}+1\right)\left(x^{6}-x^{5}+x^{2}-x\right)(\bmod 3) \text {. } \\
\end{array}
$$

Therefore, the smallest positive integer $n$ that satisfies the condition is $n=8$.