Problem 1135

For $n=1$ we have $27-2=25=5^{2}$. We will prove that there are no other positive integers with this property.
If $n$ is odd, $n=2k+1$, then
$$\begin{aligned} 27^{n}-2^{n} &= 27^{2k+1}-2^{2k+1} = (28-1)^{2k+1}-2 \cdot 4^{k} \\ &= 4m+(-1)^{2k+1} = 4m-1 \end{aligned}$$
It is easy to see that integers of the form $4m-1$ are not perfect squares.
If $n$ is even, $n=2k$, then $27^{n}-2^{n} = (27^{k})^{2} - 2^{2k}$, and we have
$$(27^{k}-1)^{2} < (27^{k})^{2} - 2^{2k} < 27^{k},$$
hence $27^{2k}-2^{2k}$ is between two consecutive perfect squares, that is it cannot be a perfect square. The only solution is $n=1$.

(Wael Hussain Al Saeed) Assume that $n \geq 2$. For any integer $x$, we have $x^{2} \equiv 0,1 \pmod{3}$ and $x^{2} \equiv 0,1 \pmod{4}$. Considering the relation $27^{n}-2^{n}=x^{2}$ modulo 3 we get
$$27^{n}-2^{n} \equiv (-1)^{n+1} \pmod{3}$$
hence $(-1)^{n+1} \equiv 1 \pmod{3}$. It follows $n=2k+1$ for some positive integer $k$.
Considering the relation $27^{n}-2^{n}=x^{2}$ modulo 4 we obtain
$$(-1)^{n} \equiv x^{2} \pmod{4}$$
hence $(-1)^{n} \equiv 1 \pmod{4}$, that is $n=2m$ for some positive integer $m$, not possible.