Is Fourier series always used for periodic signals and Fourier transform for aperiodic signals only?

A Fourier series is only defined for functions defined on an interval of finite length, including periodic signals, as you can see from the definition of the Fourier coefficients (in the basis $\{e^{inx}\}_{n\in\mathbb{Z}}$) $$ a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}~dx. $$

You can't define an aperiodic signal on an interval of finite length (if you try, you'll lose information about the signal), so one must use the Fourier transform for such a signal.

In principle, one could take a Fourier transform of a periodic signal in the sense that one could extend the signal outside the interval $[-\pi,\pi]$ by zero. But the resulting Fourier transform would look like $$ \widehat{f}(p) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ipx}~dx $$ and this isn't particularly different from the Fourier coefficients. Moreover, being able to express a periodic signal as a discrete sum of frequencies is a stronger statement than expressing it as a continuous sum via the inversion formula.

Edit: In response to the comment.

I mean "stronger" rather loosely, not in the mathematical sense of one statement implying another.

Taking the Fourier transform of a periodic signal by extending it to $0$ outside $[-\pi,\pi]$ gives no information that the usual Fourier transform would not. Furthermore, the inversion formula still holds, and therefore we see that $f$ can be recovered from its Fourier transform as a continuous sum over all frequencies.

What is remarkable about periodic functions is that one does not actually need all this information to recover the function; out of uncountably many values $\widehat{f}(p)$, one only needs the integer values $\widehat{f}(n)$ (i.e. the Fourier coefficients) to reconstruct the function.

So in this sense, for a periodic function, there is more to say than what the Fourier transform alone provides.