can you help me understand the field extensions by verifying/pointing out what's wrong with this argument.
Let $F_{25} = \frac{Z_5[x]}{<x^2+2>}$.
How many subfields does $F_{25}$ have?
Solution: Since char($F_{25}$) = 5, then it contains the smallest subfield is $Z_5$. Then if T is a subfield by the tower law we have $2 = |F_{25}:Z_5|=|F_{25}:T||T:Z_5|$ from which we get that $|F_{25}:T|=1,2$.
If it equals $1$, then we have $1$ vector and |T| scalars and we want to be able to generate $F_{25}$ $\Rightarrow$ $T=Z_{25}$
If it equals $2$, then similarly we get $T=Z_5$.
Is that true?
Yes this is correct. You've correctly identified that a subfield of $\mathbf{F}_{5^2}$ must contain the prime field $\mathbf{F}_5$ and hence is an intermediate field of the extension $\mathbf{F}_{5^2}/\mathbf{F}_5$ and so must have degree $1$ or $2$ over $\mathbf{F}_5$ and equally, $\mathbf{F}_{5^2}$ must have degree $2$ or $1$ over $T$.
I would prefer to argue that $|T| = 5^{[T : \mathbf{F}_5]}$ so either $T$ has 5 elements ($T = \mathbf{F}_5$) or 25 elements ($T = \mathbf{F}_{5^2}$) but the idea you have is correct as well. However:
you do not have $1$ vector, you have $|T|$ vectors, it is the dimension of $\mathbf{F}_{25}/T$ which is $1$
"we want to be able to generate $\mathbf{F}_{25}$" does not explain well why $T = \mathbf{F}_{25}$