Let F be a field of order 32. Show that the only subfields of F are F itself and {0,1}.

More generally, a field with $p^n$ elements contains a subfield with $p^m$ elements iff $m$ divides $n$.

In your case, we have $p=2$ and $n=5$, which has no nontrivial divisors.

Here is a proof of one direction, the one that concerns the question:

If a field $F$ has $p^n$ elements and contains a subfield $K$ with $p^m$ elements, then $F$ is a finite dimensional vector space over $K$ and so $p^n=(p^m)^d=p^{md}$, where $d$ is the dimension of $F$ over $K$.

Hint: if $K$ is a subfield of $F$, then in particular, $K^*$, the multiplicative group of $K$ must be a subgroup of $F^*$