If $x^2 + y^2 - 3 = 0$ has no rational solutions, then $x^2 + y^2 - 3^k = 0$ has no rational solutions. 

  
*Explain why $x^2 + y^2 – 3 = 0$ not having any rational solutions (Exercise 20) implies $x^2 + y^2 – 3^k = 0$ has no rational solutions for $k$ an odd, positive integer. (Book of Proof by Hammack)
  

Please critique my proof:
For the sake of contradiction, suppose that $x^2 + y^2 – 3 = 0$ has no rational solutions but $x^2 + y^2 – 3^k = 0$ has rational solutions for $k = 2j + 1$, 
$j\in\mathbb{Z}$. It follows that there are integers $a$, $b$, $c$ and $d$, with $c ≠ 0$ 
and $d ≠ 0$, such that $x = a/c$ and $y = b/d$ are solutions of $x^2 + y^2 – 3^k = 0$. 
Now, we have that $a^2/c^2 + b^2/d^2 = 3^{2k} \cdot 3^1$, it follows                  that $a^2/(3^{2k}c^2) + b^2/(3^{2k}d^2) = 3$. This in turn leads to that there are some integers $e = 3kc$ and $f =3kd$ such that $x = a/e$ and $y = b/f$ are rational solutions of $x^2 + y^2 = 3$, a contradiction.

Essentially correct, but more complicated than required. You make some confusion about $k$ and $j$, though.
Suppose $x$ and $y$ are rational solutions of $x^2+y^2=3^k$, with $k$ an odd integer. Then $k=2j+1$ and so
$$
\left(\frac{x}{3^j}\right)^{\!2}+\left(\frac{y}{3^j}\right)^{\!2}=3
$$
where $x/3^j$ and $y/3^j$ are rational.
