This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$
$1^3 + 2^3 + ... + (k+1)^3$
$ = 1^3 + 2^3 + ... + k^3 + (k+1)^3$
$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.
Annnnd I'm stuck. Not sure how to proceed from here on.
HINT: You want that last expression to turn out to be $\big(1+2+\ldots+k+(k+1)\big)^2$, so you want $(k+1)^3$ to be equal to the difference
$$\big(1+2+\ldots+k+(k+1)\big)^2-(1+2+\ldots+k)^2\;.$$
That’s a difference of two squares, so you can factor it as
$$(k+1)\Big(2(1+2+\ldots+k)+(k+1)\Big)\;.\tag{1}$$
To show that $(1)$ is just a fancy way of writing $(k+1)^3$, you need to show that
$$2(1+2+\ldots+k)+(k+1)=(k+1)^2\;.$$
Do you know a simpler expression for $1+2+\ldots+k$?
(Once you get the computational details worked out, you can arrange them more neatly than this; I wrote this specifically to suggest a way to proceed from where you got stuck.)
Consider the case where $n = 1$. We have $1^3 = 1^2$. Now suppose $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$. Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$. Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$. Hence, the statement holds for the $n + 1$ case. Thus by the principle of mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
IMHO, this fact is a coincidence; a better approach is to prove the closed-form formula for both. As we know
$$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $$
the corresponding claim to prove is
$$ 1^3 + 2^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4} $$