Why isn't this undecidable problem in NP?
I think you misunderstood what it means to solve a diophantine equation, and Matiyasevich's indecidability theorem.

Matiyasevich proved that for every RE set $S$ there is a diophentine equation $f(n;x_1,...,x_k)$ such that $n \in S$ only if there exists integer coefficients $x_1$,..,$x_k$ such that $f(n;x_1,...,x_k) = 0$. In particular, the halting problem is a typical RE set, and so solving the above problem is undecidable. 

Note that $x_1, ... x_k$ are not bounded in size, and in general can be arbitrarily large, so there is no "polynomial sized certificate" evident in this problem.

To expand: the integers $x_1, ... , x_k$ need to be written in binary to be a certificate. Since these integers can be arbitrarily large (regardless of $n$), we have that the certificate is not polynomial in $\log n$ or more importantly, not bounded by and computable function.

Every problem in $\mathsf{NP}$ however, has a certificate that is bounded by some polynomial $p(N)$ (where $N$ is size of input). So $\mathsf{NP}$ questions are trivially decidable, since you can enumerate every possible bit string upto length $p(N)$ and if none of them certify the input, return false. If some does then return true.