Graphs with only disjoint perfect matchings

$m=3$ is indeed the maximum, and $K_4$ is the only example for this value of $m$.

Two perfect matchings form a disjoint union of cycles. If there is more than one cycle, then you may swap one of them, obtaining a third matching on the same edges. So any two of the $m$ matchings form a Hamiltonian cycle.

Assume that $m\geq3$; consider a Hamiltonian cycle $v_1,\dots,v_{2n}$ formed by the first two matchings, and check how the third one looks like.

If some its edge $(v_i,v_j)$ subtends an arc of odd length (i.e. if $i-j$ is odd), then we may split the vertices outside this arc into pairs, and split the cycle $v_i,\dots,v_j$ into edges including $(v_i,v_j)$, obtaining a matching intersecting the third one but not coinciding with it. This should not be possible; thus each subtended arc is even.

Now take an edge $(v_i,v_j)$ subtending minimal such arc, and consider an edge $(v_{i+1},v_k)$ of the third matching (going otside this arc). Now split the cycle $v_i,v_j,v_{j+1},\dots,v_k,v_{i+1}$ into edges containing $(v_i,v_j)$, and split all the remaining vertices into edges of the Hamiltonian cycle (it is possible, according to the parity). If $2n>4$, you again get a fourth matching sharing edges with the third ones bot diferent from it.

Thus $2n\leq 4$ and $m\leq 3$.

It looks like Ilya Bogdanov has answered your question, but here is a pointer to related literature.

An edge that appears in at most one perfect matching is known as a forcing edge. Che and Chen have written a survey of the literature on forcing edges and related topics.