Example 32 (2004 Slovenia National Team Selection Test) Find all positive integers $n$, such that $n \cdot 2^{n-1} +$ 1 is a perfect square.
Let $n \cdot 2^{n-1} + 1 = m^2$, i.e.,
$$
n \cdot 2^{n-1} = (m+1)(m-1).
$$

It is evident that $n=1, n=2, n=3$ do not satisfy the condition.
Assume $n > 3$, then the value on the left side of equation (1) is even, hence $m$ must be odd.
Let $m = 2k + 1$, thus we have $n \cdot 2^{n-3} = k(k+1)$.
Since the consecutive integers $k$ and $k+1$ are coprime, $2^{n-3}$ must exactly divide one of $k$ and $k+1$, which implies $2^{n-3} \leq k+1$.
This leads to
$n \geq k, 2^{n-3} \leq n+1$.
The latter inequality holds for $n=4$ and $n=5$.
Using mathematical induction, it can be easily proven that for $n \geq 6, 2^{n-3} > n+1$.

Upon inspection, $4 \times 2^3 + 1 = 33$ is not a perfect square, while $5 \times 2^4 + 1 = 81$ is a perfect square.