From your reasoning, we know that $N$ is $1$ or $9$. If it is $1$, then $U$ is $9$, and if it is $9$, $U$ is $0$, because $U$ is a result of the $E-E$ with possible borrowing from $SEVEN - THREE$. This means no other letter can be $9$.
We know that $S = T+1$ and $H > E$, because there had to be a borrowing to get a four-digit number. Since none of these letters can be $9$, we know that $E$ and $T$ also can't be $8$. The condition that $TEN$ is a square, leaves $169,361,529,729$. However, in the last case, $R = N-E$ would be $7$, not distinct from $T$, and in the second case, $R = 10 + N - E = 5$, making $FOUR$ divisible by $5$. So $N = 9$, $U = 0$, and $TEN = 169,529$. This means $SEVEN$ is either $26V69$ or $62V29$. However, in the first case, that would mean $16 - H = F$, but that's not possible as they must be distinct and neither can be $9$ (there can't be a borrow because $R = 3$ and $1$ and $2$ are already taken as values for $V$). So $SEVEN = 62V29$.
So $SEVEN - THREE = FOUR$ is now $62V29 - 5H722 = FO07$. We know $F + H = 11$ or $F + H = 12$, depending on whether there's a borrow. Since $RUOF$ is prime and the only odd digits left are $1$ and $3$, $F$ must be one of them, but it can't be $1$ and satisfy either equation. So $F$ is $3$. $H$ must then be $8$, because $9$ is taken. Thus, there must have been a borrow in $V - 7 = O$, which is satisfied by $V = 1$, $O = 4$.
Thus, $SEVEN - THREE = FOUR$ is
62129 - 58722 = 3407
and checking the prime table shows $62129$, $3407$, and $7043$ are indeed prime. Meanwhile, $TEN = 529 = 23^2$ is a perfect square, as desired.
As a fun aside, $FOURTEEN$ is prime but $SEVENTEEN$ is not.
EDIT: removed reliance on prime table.