Given coprime $p,q=\Theta(\ell)$ and integer $t\geq3$ is there coprime $m,n=\Theta(\ell^{t-1})$ with $(mn,pq)=1$ such that there are $\alpha_i\in\Bbb Z$ and $m'\in\Bbb Z$ with
$$\alpha_1p^{t-2}q^t+1=m'm^2$$
$$\alpha_2p^{t-1}q^{t-1}+1=m'mn$$
$$\alpha_3p^tq^{t-2}+1=m'n^2$$
where $m,n$ where $0<m'<p^{t}q^t$ holds?

There seems to be room for such $m',m,n,\alpha_i$. Such an $m'$ has to be of form $p^2q^2m''$ where $0<m''<p^{t-2}q^{t-2}$ holds.

$\alpha_i=O(\ell^{2t})$ holds.

The expected intersection of choices is $$\frac{\mbox{# choices of }\alpha_i\times\mbox{# choices of }m',m\mbox{ and }n}{(\mbox{# of choices of resiudes modulo }p^{t-2}q^{t-2})^3}\approx\frac{\ell^{6t}\times \ell^{2t-4}\ell^{t-1}\ell^{t-1}}{\ell^{6t-6}}=\ell^{4t}\gg1.$$

A more pessimistic argument (possibly realistic) is as follows. For fixed $m'$ roughly $\Theta(\ell^{t-1})$ values ($-1$ is exponent because $m'$ is bigger than $p^{t-2}q^t$ by $q^2$) of $\alpha_1$ hit $m'$ multiple of a square and roughly $\Theta(\ell^{t-1})$ values ($-1$ is exponent because $m'$ is bigger than $p^{t}q^{t-2}$ by $p^2$) of $\alpha_3$ hit $m'$ multiple of a square. Total possible values of that $\alpha_2p^{t-1}q^{t-1}+1$ can hit is $\Theta(\ell^{2t})$. On other hand size of $m'mn$, $m'm^2$ or $m'n^2$ is $\Theta(\ell^{t-2}\ell^{t-2}\ell^{t-1}\ell^{t-1})=\Theta(\ell^{4t-6})$. So we can expect $\Theta(\frac{\ell^{t-1}\ell^{t-1}\ell^{2t}}{\ell^{4t-6}})=\Theta(\ell^4)$. So for every fixed $m'$ there should be $\Theta(\ell^4)$ triples of $(\alpha_1,\alpha_2,\alpha_3)$ that produce multiple $m'$ of some squares $m^2,n^2$ with some $\alpha_2p^{t-1}q^{t-1}+1$ colliding at multiple $m'$ of $mn$.