Here is one way to proceed:
Suppose $m^3=n^2$, and $n$ is even. As you say, $n$ must be $2k$ for some integer $k$. Then $m^3=4k^2$, which means in particular that $2\mid m^3$. Now, either by a quick contradiction, or by using Euclid's lemma, it follows that $2\mid m$. (The point here is that $2$ is prime, so if $2\mid m^3=m\cdot m\cdot m$, then it must divide $m$ or $m$ or $m$. The argument by contradiction is that if $m$ is odd, then $m^3$ is also odd by a direct computation, so $2\mathrel{\not|} m^3$.)
If $2\mid m$ then there is an integer (say, $\ell$) such that $m=2\ell$. Then $m^3=8\ell^3$. Then $8\ell^3=4k^2$, so $2\ell^3=k^2$, so $2\mid k$. Then $k=2a$ for some integer $a$, and $n=2k=4a$ is indeed a multiple of $4$.
But it may be worth thinking about this in slightly more general terms. The equation $m^3=n^2$ tells us that the prime factorizations of the two numbers $m^3$ and $n^2$ must coincide. Start with the prime factorization of $m$ and cube it to find the prime factorization of $m^3$. Similarly, start with the prime factorization of $n$ and square it to find the prime factorization of $n^2$. This tells you that there are primes $p_1 and $q_1 and exponents $\alpha_1,\dots,\alpha_a$ and $\beta_1,\dots,\beta_b$ such that $ p_1^{3\alpha_1}\dots p_a^{3\alpha_a}=q_1^{2\beta_1}\dots q_b^{2\beta_b}. $ (Here, of course, $m=p_1^{\alpha_1}\dots p_a^{\alpha_a}$ and $n=q_1^{\beta_1}\dots q_b^{\beta_b}$.)
The uniqueness of the prime factorization tells us that $a=b$, $p_1=q_1$, $p_2=q_2$, etc, and $3\alpha_1=2\beta_1$, $3\alpha_2=2\beta_2$, etc.
Now: If $3\alpha_i=2\beta_i$, then $2\mid\alpha_i$ (so $m$ is a square) and $3\mid\beta_i$ (so $n$ is a cube), so $n$, being even, must in fact be a multiple of $2^3=8$.
The advantage of this approach, which perhaps looks a bit more complicated than the first one, is that it generalizes to other situations. So one can now quickly conclude by the same argument that if, say, $m^6=n^{17}$, then $n$ is a sixth power (so if $n$ is even, then it is a multiple of $64$, for example). Or, if $m^6=n^{15}$, then $n$ is a square. Etc.