Pumping lemma on {a^n | n=3^k} — help finishing the proof
It is not enough. Also note that you didn't define $x$ correctly, and forgot $z$.
I would define:
$$x = a^s$$
$$y = a^t$$
$$z = a^{3^m - s -t}$$
such that
$s + t &lt;= m$ and $t &gt;= 1$ (hence $1 &lt;= t &lt;= m$)

You need to show that for $i=2$, you get something that is not in L (meaning, there is no $k$ such that $x(y^2)z = a^{3^k}$ ).

Let's look at $x(y^2)z = a^{3^m+t}$. 
Its length is $3^m + t$.

Using the fact that for every $i, i &lt; 3^i$, We get:
$3^m &lt; 3^m + 1 \le |x(y^2)z| = 3^m + t \le 3^m + m &lt; 3^m + 3^m = 3^{m+1}$

We see that the length of $xy^2z$ isn't in the form of $3^n$ (it's something between $3^m$ to $3^{m+1}$, hence it is not in L. 
Contradiction to the pumping lemma.