Make $2^8 + 2^{11} + 2^n$ a perfect square

Hint: $(2^a + 2^b)^2 = 2^{2a} + 2^{2b} + 2^{a+b+1}$.

If $0\le n\le 7$, then there are no solutions. Let $n\ge 8$.

$$2^8+2^{11}+2^n=\left(2^4\right)^2\left(9+2^{n-8}\right)$$

is a square if and only if $9+2^{n-8}=m^2$ for some $m>3$, i.e. $2^{n-8}=(m+3)(m-3)$, so $m+3=2^k$ and $m-3=2^l$ for some $k>l\ge 0$. If $k\ge 4$, then $$6=2^k-2^l\ge 2^k-2^{k-1}\ge 8>6$$ contradiction, so $k\in\{1,2,3\}$, which only gives $k=3$, so $m=5$, $n=12$.

$2^8 + 2^{11} + 2^n = 2^8(1 + 8 + 2^{n-8})=2^8(9 + 2^{n-8})$

Therefore, $9 + 2^{n-8}$ has to be a perfect square. Clearly, $9 + 16 = 25$ is a perfect square.

So, $2^{n-8} = 2^4$ giving, $$n = 12$$