Find all integers $n$ such that $2^{n}+3$ is a perfect square. The same question with $2^{n}+1$.
A square is never congruent to 3 modulo 4 (if you don't believe me, try it and see!), so $2^{n}+3$ cannot be a square if $n \geq 2$. And if $n=1$ then $2^{1}+3=5$ is not a square and if $n=0$ then $2^{0}+2=3$ is not a square. The second question is less easy. We are looking for $n$ and $x$, two integers, such that $2^{n}+1=x^{2}$. In other words:

$$
2^{n}=x^{2}-1=(x-1)(x+1)
$$

so $x+1$ and $x-1$ are both powers of 2. The only powers of 2 that are 2 units apart are 2 and 4, so $2^{n}=(x-1)(x+1)=2 \cdot 4=8$ thus $n=3$ is the only solution.