Say $27^{n}-2^{n}=x^{2}$, obviously $x$ is odd For $n$ even, let $n=2k$ for a natural number $k$. Then $(27^{k})^2-x^2=(27^{k}-x)(27^{k}+x)=2^{2k}$ Notice that $gcd(27^{k}-x,27^{k}+x)|2^{2k}$, so $gcd(27^{k}-x,27^{k}+x)=2^a$ for a natural number $a$(since $x$ is odd) $gcd(27^{k}-x,27^{k}+x)=gcd(27^{k}-x,2x)=2.gcd((27^{k}-x)/2,x)=2^{a} \rightarrow a=1, gcd(27^{k}-x,27^{k}+x)=2$ So, since $27^{k}-x<27^{k}+x$, then $27^{k}-x=2$ and $27^{k}+x=2^{2k-1}$ which results in $27^{k}=2^{2k-2}+1$ But this leads to $2^{2k-2}=27^{k}-1=26.(27^{k-1}+27^{k-2}+...+1) \rightarrow 26 \mid 2^{2k-2}$, but $13 \nmid 2^{2k-2}$, contradiction So, for $n$ odd, if $n>1$, then $2^{n} \equiv 0(mod 4)$, so $x^2=27^{n}-2^{n} \equiv 3^{n} (mod 4) \equiv 3(mod 4)$. which does not satisfies For $n=1, 27^{1}-2^{1}=25=5^{2}$, which satisfies, so $n=1$ is the only solution