Finding Pareto efficient allocations via social planner's problems is a special case of scalarization in convex optimization.
Suppose agents' utility functions are $u_i:\mathbb{R}^n \rightarrow \mathbb{R}$, for $i = 1, 2, \cdots, m$, and the feasible allocations are given by $g(x) \geq 0$ for some $g:\mathbb{R}^n \rightarrow \mathbb{R}^p$.
The general result is as follows.
Sufficiency under general conditions
If an allocation $x$ solves the social planner's problem with strictly positive social weights, then it is Pareto. This is true with no assumptions on utility functions and feasibility constraint.
In other words, if
$$
x \in \arg\max_{g(x') \geq 0} \sum_{i = 1}^m \lambda_i u_i(x'), \mbox{ for some } \lambda \in \mathbb{R}^m_{++}, \quad (*)
$$
then it is Pareto optimal.
This is easy to see. If $y$ Pareto-improves $x$, then
$$
\sum_{i = 1}^m \lambda_i u_i(y) > \sum_{i = 1}^m \lambda_i u_i(x)
$$
for any $\lambda \in \mathbb{R}^m_{++}$. It is also easy to see that zero entry in $\lambda$ cannot be allowed---it would make the statement false.
The converse is not true in general---this sufficient condition is not necessary. There are Pareto allocations that do not solve $(*)$. This is because the set of achievable utility values
$$
\mathcal{U} = \{ (u_1 (x), \cdots, u_m(x)); g(x) \geq 0 \}
$$
is in general not a convex subset of $\mathbb{R}^m$. The social planner's weights $\lambda$ corresponds to certain supporting hyperplanes of $\mathcal{U}$. If $\mathcal{U}$ is not convex, its Pareto frontier cannot be recovered by varying $\lambda$.
However, there is a partial converse.
Necessity under concavity
Assume $u_i$'s and $g$ are concave. If an allocation $x$ is Pareto, then it solves a social planner's problem with nonnegative social weights.
In other words, if $x$ is Pareto then
$$
x \in \arg\max_{g(x') \geq 0} \sum_{i = 1}^m \lambda_i u_i(x') \mbox{ for some } \lambda \in \mathbb{R}^m_{+}. \quad (**)
$$
This is true because of convex sets and separating hyperplanes.
This is a partial converse because it's necessary but not sufficient. There are allocations that solve $(**)$ but not Pareto.
Comment
Even under concavity, $(**)$ is a only a partial converse of $(*)$. Notice the gap between strictly positive and nonnegative social weights.
To recover the Pareto frontier, the general procedure is to first find solutions to $(*)$. Then check the solutions to $(**)$ case-by-case for Pareto optimality. Alternatively, one can also take the closure (the limit points) of solutions to $(*)$.
