I will give a few references which could be considered as some answer to this question. Though the matter of "straight forward" is up to interpretation. Furthermore, I am far from any kind of expert on these sequences. I have written a paper on (some combinatorial aspects of) Lucas sequences and remember some references from when I was reading up on them.
First there are a lot of second order linear reccurences which have few (= finitely many) perfect square terms. In Perfect powers in second order linear recurrences by Pethö a special case of the main theorem says that for a second order linear recurrence $G_n = AG_{n-1} - BG_{n-2}$ there are only finitely many perfect square terms provided some conditions. The conditions are that $A \neq 0$, $|G_0| + |G_1| \neq 0$, $\gcd(A,B) = 1$, $A^2 \neq iB$ for $i \in \{1,2,3,4\}$, and $D$ is not a perfect square if $BC = 0$. Here $C = G_1^2 - AG_0G_1 + BG_1^2$ and $D = A^2 - 4B$. This paper gives some bounds on where perfect squares can arise by Baker’s method, but since Lucas sequences are special one may want to know more.
In The Square Terms in Lucas Sequences by Ribenboim and McDaniel is in shown more precisely where perfect square terms can arise in Lucas sequences under some conditions. Under the conditions that $P$ and $Q$ are odd, $\gcd(P,Q) = 1$, and $D = P^2 - 4Q > 0$ it is shown that
- if $V_n(P,Q)$ is a perfect square, then $n \in \{1,3,5\}$.
- if $U_n(P,Q)$ is a perfect square, then $n \in \{0,1,2,3,6,12\}$.
Here the notation matches with the usual notation which is given the OEIS link "Lucas sequences" in the question. We see this result agrees with what is known for $F_n$ and $L_n$. Again I am not an expert, but I think exactly where squares occur depends on particular choices of $P$ and $Q$ and I am not familiar with particular of methods used in various cases. You can find more papers like Lucas sequences whose 12th or 9th term is a square
by Bremner and Tzanakis. Looking at these papers with their references + google scholar gives more information on this problem.