I'm trying to prove that for two modular forms $f$ and $g$ of weight $k$ and $k'$ respectively, that are congruent modulo a prime $\ell\ge 5$, their weights are congruent modulo $\ell-1$. This is what I have so far: if $f\equiv g\pmod\ell$ then $f\left(\frac{az+b}{cz+d}\right)\equiv g\left(\frac{az+b}{cz+d}\right)\pmod\ell\implies (cz+d)^{k}f(z)=(cz+d)^{k'}g(z)\pmod\ell$. \So$(cz+d)^{k}\equiv (cz+d)^{k}\pmod\ell$. So $(cz+d)^{k-k'}\equiv 1\pmod\ell$. If $c\equiv 0\pmod\ell$ then $d^{k-k'}\equiv 1\pmod\ell$ and by Fermat's little theorem $d^{k-k'}\equiv d^{\ell-1}\pmod\ell$. So $k-k'\equiv 0\pmod{\ell-1}\implies k\equiv k'\pmod{\ell-1}$
Is this the right way to go and if so is it possible to extend this for all $c$? Thank you.
Why don't you read one of the papers of Serre where this result is proved? The argument is not too difficult if you know some theory, but is not something you will be able to work out for yourself if all you know is the bare definition of a modular form. (Serre's Antwerp article in LNM 350 proves it, and it might also be proved in his Bourbaki seminar "apres Swinnerton-Dyer".)
As David Loeffler commented, your current argument is nonsense.