Constant case CRT: $\,x\equiv a\pmod{\! 2},\ x\equiv a\pmod{\! 5}\iff x\equiv a\pmod{\!10}$

Your proof is correct. It invokes a simple special case of CRT = Chinese Remainder Theorem when the values $\,a_1 = a_2\,$ are constant, say $\,a,\,$ which is equivalent to the following basic result

UL = Universal property of LCM: $\ \rm \,\ j,k\mid n\!\!\color{#0a0}{\overset{\rm UL\!\!}\iff} {\rm lcm}(j,k)\mid n$

CCRT = Constant case CRT $\ $ If $\rm \,a,p,q\,$ are integers and $\rm \,\gcd(p,q) = 1\,$ then

$$\begin{align}\rm x\equiv a\!\!\pmod{p}\\ \rm x\equiv a\!\!\pmod{q}\end{align}\iff\,\rm x\equiv a\!\!\pmod{pq}\qquad$$

Proof $\ $ Below I sketch the key ideas in four proofs.

$\rm(1)\ \ \ x \equiv a\pmod {pq}\:$ is clearly a solution, and the solution is $\color{#C00}{\textit{unique}}$ $\!\!\pmod{\rm\!pq}\,$ by CRT.

$\rm(2)\ \ \ p,q\:|\:x\!-\!a\!\!\color{#0a0}{\overset{\rm UL\!\!}\iff} lcm(p,q)\:|\:x\!-\!a.\:$ Further $\rm\:\gcd(p,q)=1\!\iff\!lcm(p,q) = pq.$

$(3)\ \, $ By Euclid's Lemma: $\rm\:(p,q)=1,\,\ p\mid nq\! =\!x\!-\!a\:\Rightarrow\:p\:|\:n\:\Rightarrow\:pq\:|\:nq = x\!-\!a.$

$\rm(4)\ \, $ The list of prime factors of $\rm\,p\,$ occurs in one factorization of $\,\rm x-a\,$, and the disjoint list of prime factors of $\rm\,q\,$ occurs in another. By $\color{#C00}{uniqueness}$, the prime factorizations are the same up to order, so the concatenation of these disjoint lists of primes occurs in $\rm\,x-a,\,$ hence $\rm\,pq\mid x-a$.

Remark $\ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.

Quite frequently, $\color{#C00}{\textit{uniqueness}}\ \textit{theorems}\,$ provide powerful tools for proving equalities.

Yours is a valid, clean argument. It is based on this:

If $m$ and $n$ divide $a$, then $lcm(m,n)$ divides $a$.

In your case, you have that $2$ and $5$ divide $3^4-1$, and so $10=lcm(2,5)$ divides $3^4-1$.