(b) If $n>1$, then $4$ divides $x^2+1$, a contradiction since $x^2\equiv0,1\pmod4$. So, $n=1$ and $x=1$. (a) Re-write as $2^n=(x+1)(x-1)$. Thus $x+1=2^a,x-1=2^b$ with $a+b=n$ since there are no prime factors other than $2$. We get $2^a-2^b=2$. Therefore, $b\neq0$ and since $a>b>0$ if $a=b+c$, we have $2^b(2^c-1)=2$. This forces $b=1$ and $2^c-1=1$ or $c=1$. Thus, $a=2,b=1$ and $n=3,x=3$. There is no need to use LTE or Zsigmondy here.

@ngv, your case a and b are switched.