Budget Constraint in Utility Maximisation Problem with Lagrange Multipliers
Objective function of Lagrangian can be set up either with $+\lambda$ or $-\lambda$, depending on how you solve the budget constraint.

Actually, for the solution it does not matter if $\lambda$ has negative or positive sign in the equation. You can clearly see it from the formula if you expand the second term:
$$
\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p_{x}x - p_{y}{y})= U(x,y) +\lambda m -  \lambda p_{x}x - \lambda p_{y}{y}$$

Using the alternative approach you get exactly the same result as:

$$
\mathcal{L}(x,y, \lambda) = U(x,y) - \lambda(p_{x}x + p_{y}{y}-m)= U(x,y) -  \lambda p_{x}x - \lambda p_{y}{y} +\lambda m$$

So the resulting equations are 100% equivalent.

However, it’s easy to confuse when to use which sign so many people just memorize one of the two ways of solving Lagrangian and never use the other, because it’s easier to just memorize one of the ways than every single time thinking about the sign of lambda. There is really nothing deeper about it.