If $m^3 = n^2$ and $n$ is even, then n is divisible by $4$. Conjecture: If $m^3 = n^2$ and $n$ is even, then n is divisible by $4$.
The proof falls apart from the beginning.
$n$ is even therefore there is a number $k$ such that $n=2k$
$m^3 = n^2$
$m^3 = (2k)^2$
$m^3 = 4k^2$
$4|4k^2$ therefore $4|n^2$ 

However, I can't think of an example where a cubic is equal to a square.  I also ask with hesitation because we have been studying prime numbers and the Euclidean Key theorem as well as other proofs using the Fundamental Theorem of Arithmetic.  So, this approach seems out of place for the section of homework that I'm doing.    
 A: Since $n^2 = m^3$ (i.e. $n^2$ is both a perfect square and a perfect cube), the number of factors of 2 in its prime factorization must be a common multiple of 2 and 3.
The smallest such multiple would be zero, but we can discard that possibility on the grounds that $n$ is even.
So, $n^2$ has at least six factors of 2 in its prime factorization. This means that $n$ has at least three factors of 2 in its prime factorization, and so $n$ is divisible by 4 (indeed, it is divisible by 8).
A: There are multiple problems here:
(1) If $n$ is even then $4|n^2$, but not necessarily $n$.
(2) Your "proof" doesn't really use the fact that $m^3 = n^2$.
Also, there are numbers that satisfy $m^3 = n^2$.  For instance, $2^6 = 4^3 = 8^2$.
A: 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 $p_1<p_2<\dots<p_a$ and $q_1<q_2<\dots<q_b$ 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.
A: First try to prove that every number that is both a square and a cube is a sixth power.  (It is easier to prove that every sixth power is both a square and a cube.)  Then you have $m^3=n^2=\ell^6$.  If $4\mid \ell^6$ then $2\mid \ell^3$.  Then you can show that $2\mid\ell^3$ only if $2\mid\ell$.  That implies that $8\mid\ell^3=n$, so of course $4\mid n$.
