Estimate complexity of integer sequence enumeration
Here is a simple recursive solution that runs in exponential time.

Geometrically you want to enumerate the lattice points which lies inside
the first quadrant of the $n$-hypercube of height $m$ and on the intersection of the $D_1$-radius $L_1$ sphere and the $D_2$-radius $L_2$ sphere (let call them respectively $S_1$ and $S_2$), up to reordering of the coefficients.

Denoting by $V_n$ the volume of the $n$  unit dimensional ball, we have :


$m^n/n!$ points in the intersection of the hypercube and the quadrant (let's call it $D$) with the ordering ,
$\alpha = O\left(m^n e^{\frac{-(D_1-nm/6)^2}{3nm}}\right)$ in $S_1\cap D$ (see theorem 1.3 of http://www.doiserbia.nb.rs/img/doi/1452-8630/2008/1452-86300802222R.pdf), considering the ordering condition.
$\beta = \frac{1}{n!2^{n}}V_n D_2^{n/2} + O\left(D_2^{n/2-1+\epsilon}\right)$ in $S_2$ (Generalized Gauss circle problem+quadrant condition+order condition). I'm not considering intersection with the hypercube here, so the global complexity will be an overestimation.


Depending on your parameters choose the minimum of $\alpha$ or $\beta$.
If its $\alpha$ perform a recursive enumeration of the integers partitions of length $n$ with the restriction on $m$ and use early abort to 
discard solutions which are getting out $S_2$.
In the other case, enumerate the lattice point on $S_2$ ( with recursive orthogonal projections for instance) and check for the appartenance in $S_1$ for each returned vector.

Hence the global complexity will be dominated by $min(\alpha, \beta)$.