I was wondering if there were any two integers $a$ and $b$ where $a^3=b^2$.
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Yes. $a=n^2$ and $b=n^3$, where $n$ is any integer. (for example, $n=2$ yields, $a=4$ and $b=8$).
It is actually easy to prove using the Fundamental Theorem of Arithmetic that these are all solutions.
P.S. I am really surprised that you missed the obvious solutions: $a=b=0$ and $a=b=1$....
Hint $\rm\ a=0\!\iff\! b=0.\:$ Else $\rm\:(b/a)^2 = a\in\Bbb Z,\:$ so $\rm\:b/a = n\in\Bbb Z$ via Rational Root Test (RRT). Therefore $\rm\ a = (b/a)^2 = n^2,\:$ so $\rm\:b = an = n^3,\:$ and, indeed, $\rm\:a^3 = (n^2)^3 = (n^3)^2 = b^2\ \ $ QED
Remark $\ $ Note that the proof did not require unique factorization but only the much weaker Rational Root Test, monic-case. Thus the proof generalizes to any integrally-closed domain.
Update (to answer questions in comments) Suppose that $\rm\:x^2 - a\:$ has a rational root $\rm\:x = b/a.\:$ Cancelling $\rm\:gcd(a,b)\:$ we can write $\rm\:x = c/d\:$ in lowest terms. Then RRT implies that the denominator divides the lead coef, i.e. $\rm\:d\:|\:1,\:$ so $\rm\:d=\pm1,\:$ so $\rm\: x = c/d = \pm\, c\in \Bbb Z,\:$ hence $\rm\:b/a = x = c/d\in\Bbb Z.$
Any number in the form of $n^6$ can be expressed in the desired form so there will be infinite solutions for $a$ and $b$.