The two solutions are
$179 \times 179 = 32041$
$189 \times 189 = 35721$
Proof
First of all given that $\sqrt{100000} < 317$, we find that $TWO < 317$ and, in particular, $T$ must be $1,2$ or $3$ (assuming no leading zeroes).
Also, since $T$ is the remainder of a square number divided by $10$, it can only be $0,1,4,5,6,9$ so it must be that $T=1$.
Since $O^2$ must end in $1$, it can only be $1$ or $9$ and since $O\neq T$, it must be that $O=9$.
This actually just leaves us with $8$ options to try ($W=0,2,3,4,5,6,7,8$) so we can test them all by hand and it only requires that the remaining unidentified are distinct and different from $1$ and $9$.
Also, since $E>1$, we only have to try $W=4,5,6,7,8$ as $\sqrt{20000} > 140$.
$149 \times 149 = 22201$ which has double $2$s so this doesn't work.
$159 \times 159 = 25281$ which also has double $2$s.
$169 \times 169 = 28561$ which puts $W=H$ so this doesn't work.
$179 \times 179 = 32041$ and this works!
$189 \times 189 = 35721$ and this also works!