Clearly : $n=1,2$ works For : $n\ge 3$ We have : $4^n<4^n+4n+1< (2^n+1)^2$ The first one is trivial for the second one , when simplifying we get : $4n<2^{n+1}$ Proof by induction : For : $n=3,12<16$ Suppose the property is true for n , let's prove it for n+1: $4n<2^{n+1}\implies 4(n+1)<2(2^n+2)$ And we have that : $2^n>2\implies 2^n\times 2-2^n>2\implies 2^n\times 2>2^n+2\implies 2^n\times 4>2(2^n+2)\implies 2^{n+1}>2(2^n+2)$ Thus : $2^{n+2}>4(n+1)$ So the only solutions are $n=1,2$

KAME06 wrote: Find all positive integers $n$ such that $4^n + 4n + 1$ is a perfect square. Generalization: Find all positive integers $n$ such that $2^n + 2n + 1$ is a perfect square. Clearly $n\in\{2,4\}$work and $n\in\{1,3\}$ not work. For $n \ge 5$, we have $2^n + 2n + 1$ is an odd perfect square. So $$2^n+2n+1\equiv 1 \pmod4.$$So $n$ is an even positive number. Let $n=2m$, where $m\ge3$. So $$\left(2^m\right)^2<2^{2m}+4m+1<\left(2^m+1\right)^2. \text{ (contradiction)}$$

Let $4^n+4n+1=a^2$. We have $a^2>4^n$ then $a>2^n$ Therfore: $4^n+4n+1=a^2 \ge (2^n+1)^2\Rightarrow n\ge2^{n-1}$ This inequality only true for $n=1;2$ and we can show these are 2 true answers.

For n>2 $$(2^n)^2<4^n+4n+1<(2^n+1)^2$$For n=1,2 works so the only solution are n=1,2