Finding the possible lengths and widths, given a surface area.

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Short Version of Question:

Each of $l$, $w$ and $k$ is a positive integer. Determine all possible values for $l$ and $w$ such that $l \ge w$, and $(k + 1)(l + w - 2k) = 133$.

Long Version of Question:

A question from the most recent of Waterloo's Hypathia contest -

A number of cubes, each with edge length $1 cm$, are arranged to form a rectangular prism having length $l cm$, widh $w cm$, and thickness $1 cm$. A frame is formed by removing a rectangular prism with thickness $1 cm$ located $k cm$ from each of the sides of the original rectangular prism, as shown. Each of $l$, $w$ and $k$ is a positive integer. If the frame has surface area $532 cm^2$, determine all possible values for $l$ and $w$ such that $l \ge w$.

And the picture that came with it is this -

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My work so far on it has been the standard math approach; to create some equations and see what happens. Here's what I got -

$Front + Back = 2lw - 2(l - 2k)(w - 2k)$

$Sides = 2l + 2w$

$Insides = 2(l - 2k) + 2(w - 2k)$

Adding all this together, we get

$2lw - 2(l - 2k)(w - 2k) + 2l + 2w + 2(l - 2k) + 2(w - 2k) = 532$

$lw - (l - 2k)(w - 2k) + l + w + l - 2k + w - 2k = 266$

$lw - (lw - 2kw - 2kl + 4k^2) + 2l + 2w - 4k = 266$

$2kw + 2kl - 4k^2 + 2l + 2w - 4k = 266$

$-2k^2 + l + w - 2k + kw + kl = 133$

$-2k(k + 1) + l(k + 1) + w(k + 1) = 133$

$(k + 1)(l + w - 2k) = 133$

And at this point, I'm stuck. I feel as though there is some strange factoring strategy that I'm missing here. Any ideas?

Thank you ahead of time.

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hint: $133=7 *19$, both primer ,LHS is two number product, can you find the solution?