When is the difference of two consecutive positive cubes a perfect square? Are there only finitely many solutions in positive integers $m,n$ to the equation $$(m+1)^3-m^3=n^2\; ?  $$
 A: There are infinitely many of them. Let 
$$f(m,n) \stackrel{def}{=} (m+1)^3 - m^3 - n^2$$
It is clear the equation $f(m,n) = 0$ has a trivial solution $(m,n) = (0,1)$.
By brute force, one can verify
$$f(7m+4n+3,12m+7n+6) = f(m,n)$$
This means if $(m,n)$ is a solution for $f(m,n) = 0$, so does $$(m',n') = (7m+4n+3,12m+7n+6).$$
Start from the trivial solution $(0,1)$, we can use this to generate infinitely many positive integer solutions for the equation $f(m,n) = 0$.
A: You have the quadratic equation $3m^2+3m+1=n^2$, which has positive discriminant. In general, every quadratic equation in two variables with a positive discriminant either has $0$ or $\infty$ solutions. Since there is at least one (namely $(m,n)=(0,1)$), there are infinitely many.
You may want to read about Pell's equation $x^2-dy^2=1$ or the more general (sometimes called Pell-type equation) $x^2-dy^2=a$ to learn why this kind of equations either has $0$ or $\infty$ solutions. In fact, by studying the Pell-type equation associated to a particular such quadratic it is possible to give explicit recursive formulas to construct solutions (see achille hui's answer here). Moreover, all solutions are given by a finite number (that can be arbitrarily large, though it's always possible to give an upper bound) of such recursive families.
A: We write the equation:
$$(x+1)^3-x^3=y^2$$
$$3x^2+3x+1=y^2$$
Let's use the formula.  https://mathoverflow.net/questions/31118/integer-polynomials-taking-square-values/195614#195614
For these equations we use the standard approach.
For a private quadratic form:  $$y^2=ax^2+bx+1$$  
Using solutions of Pell's equation:  $$p^2-as^2=1$$  
Solutions can be expressed through them is quite simple.  
$$y=p^2+bps+as^2$$  
$$x=2ps+bs^2$$  
$p,s$ - these numbers can have any sign.
Finding solutions of equations Pell - standard procedure.
So for our case. Use the Pell equation.  $p^2-3s^2=1$
And then the decision will have a look.
$$x=2ps+3s^2$$
$$y=p^2+3ps+3s^2$$
Knowing the first solution.  $( p_1 ; s_1 ) - (2 ; 1 )$
You find the following.
$$p_2=2p_1+3s_1$$
$$s_2=p_1+2s_1$$
Be aware that the number  $(p;s)$ can have any sign.
