Find all positive integers $n$ such that $2^{n}+3$ is a perfect square.
It is about solving $2^{n}+3=a^{2}$ with $a$ being a natural integer. If $2 \leq n$, then we have $a^{2} \equiv 3 \bmod 4$, which is not possible according to what we have seen earlier. Therefore, the only possibilities are $n=0$ and $n=1$, which, after verification, gives $n=0$ as the only solution.

Solution to the exercise $4 n-1 \mid 3 n+4$ so $n-1 \mid(3 n+4)-3(n-1)=7$. Since $n-1$ is positive and the only positive divisors of 7 are 1 and 7, we have $n-1=1$ or $n-1=7$, that is, $n=2$ or $n=8$. Conversely, for $n=2$, we have $1 \mid 7=3 \times 1+4$, and for $n=8$, we have $7 \mid 28=3 \times 8+4$. Solution to the exercise $5 n+2 \mid n^{2}+3 n$ so $n+2 \mid\left(n^{2}+3 n\right)-n(n+2)=n$, thus $n+2 \leq n$ that is $2 \leq 0$ which is impossible, hence there is no solution to the proposed equation.