Thanks...

I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -

\(\displaystyle log_2(log_2(log_2(x))) = 1\)

\(\displaystyle log_2(log_2(log_2(x))) = log_2(2)\) [I can write \(\displaystyle log_2(2)\) instead of \(\displaystyle 1\), can't I?]

Yes.
So,

\(\displaystyle log_2(log_2(x)) = 2\) [Now this is what I'm unsure about. Can I eliminate \(\displaystyle log_2\) from both sides in this case??]

Yes you can. This is because the function \(\displaystyle \color{red}f(x)=\log_2(x)\) is one-to-one, or injective, which you probably haven't learned yet. But it means that f(a) = f(b) implies a = b, and graphically, f(x) passes the "horizontal line test."
Next,

\(\displaystyle log_2(log_2(x)) = 2*log_2(2)\) [where \(\displaystyle log_2(2) = 1\) ]

\(\displaystyle log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4)\)

I would write \(\displaystyle \color{red}log_2(2^2)\) so people don't misinterpret as \(\displaystyle \color{red}\left(\log_2(2)\right)^2\).
\(\displaystyle log_2(x) = 4\)

\(\displaystyle \Longrightarrow log_2(x) = 4*log_2(2)\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(2)^4\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(16)\)

Thus \(\displaystyle x = 16\)

Now is it a valid method???

Yes. Good job.