This is more of a comment than an answer, but i can not add pictures in a comment.

As I mentioned earlier in the comments, an immediate upper bound to the number of such paths is $$\min_{C \subset V(G)} \alpha(C)$$ where $C$ cuts $s$ from $t$. But the following example shows that we can not always achieve this bound: ![graph][1] [1]: https://i.sstatic.net/Q0qHA.jpg
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As there are three vertex disjoint paths from $s$ to $t$, Menger's theorem says that every vertex cut contains at least three vertices. This graph without $s$ and $t$ is triangle-free, so for every cut $C$ the independence number $\alpha(C)$ is at least $2$. But we can have only one path from $s$ to $t$ as desired. (If we want two paths we can not use 2 or 5, but if we can not use them, there is only a single path.)

This is an answer that was a comment before. I left the content of the comment, because it was leading to the answer.

The problem is NP-complete. The content of the comment:

As I mentioned earlier in the comments, an immediate upper bound to the number of such paths is $$\min_{C \subset V(G)} \alpha(C)$$ where $C$ cuts $s$ from $t$. But the following example shows that we can not always achieve this bound: ![graph][1] [1]: https://i.sstatic.net/Q0qHA.jpg
.

As there are three vertex disjoint paths from $s$ to $t$, Menger's theorem says that every vertex cut contains at least three vertices. This graph without $s$ and $t$ is triangle-free, so for every cut $C$ the independence number $\alpha(C)$ is at least $2$. But we can have only one path from $s$ to $t$ as desired. (If we want two paths we can not use 2 or 5, but if we can not use them, there is only a single path.)

The proof of NP-completeness: The problem of finding an independent set of size $k$ can be reduced to this problem. Consider a graph $G$ and add two additional vertices $s$ and $t$, connect them both with every vertex of $G$. Now the maximal number of non-adjacent paths is $\alpha(G)$. Thus computing it efficiently is considered one of the hardest tasks.